GIFT  OF 
W.  G.  3.  Shonhan 


ELEMENTARY  ANALYSIS 


BY 

PEECEY  F.  SMITH,  Ph.D. 

PROFESSOR  OF  MATHEMATICS   IN  THE   SHEFFIELD   SCIENTIFIC  SCHOOL  OF 
YALE   UNIVERSITY 

AND 

WILLIAM  ANTHONY  GEANVILLE,  Ph.D. 

INSTRUCTOR  IN  MATHEMATICS  IN  THE  SHEFFIELD  SCIENTIFIC  SCHOOL  OF 
YALE  UNIVERSITY 


GINN  AND  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


Entered  at  Stationers'  Hall 


Copyright,  1910,  by 
Percey  F.  Smith  and  William  Anthony  Granville 


all  bights  reserved 
510.9 


e. 


GINN  AND  COMPANY  •  PRO- 
PRIETORS '  boston  •  U.S.A. 


PREFACE 

The  text  of  this  volume  is,  to  a  considerable  extent, 
identical  with  portions  of  corresponding  chapters  in  Smith 
and  Gale's  "Elements  of  Analytic  Geometry"  and  Gran- 
ville's "Elements  of  the  Differential  and  Integral  Calcu- 
lus." The  new  material  is  contained  in  the  chapters  on 
Curve  Plotting  (Chapter  V)  and  Functions  and  Graphs 
(Chapter  VI).  At  the  same  time,  the  parts  which  have 
appeared  in  previous  books  of  the  series  have  been  thor- 
oughly revised  and,  to  a  considerable  extent,  rewritten,  to 
the  end  that  the  aim  of  the  authors  might  be  accomplished, 
—  namely,  to  prepare  a  simple  and  direct  exposition  of 
those  portions  of  mathematics  beyond  Trigonometry  which 
are  of  importance  to  students  of  natural  science.  In  this 
connection  attention  may  be  called  to  the  intentional 
avoidance  of  anticipating  difficulties,  —  a  feature  which 
is  not  common  in  textbooks.  To  particularize,  processes 
which  are  natural  are  introduced  without  explanation,  and 
exact  definition  is  not  given  until  the  student  is  familiar 
by  practice  with  the  matter  in  hand.  Again,  in  the  deriva- 
tion of  certain  formulas  in  the  Differential  Calculus  the 
evaluation  of  particular  limits  is  not  undertaken  until  the 
student  sees  that  this  work  must  be  done  before  the  prob- 
lem can  be  solved. 

V 

401141 


vi  ELEMENTARY  ANALYSIS 

In  many  instances,  when  deemed  wise,  a  general  discus- 
sion is  introduced  by  concrete  examples.  This  feature,  so 
common  in  school  texts,  is  strangely  absent  from  books  in- 
tended for  use  in  colleges  and  technical  schools.  Interest  in 
the  subject  is  usually  aroused  in  this  way,  and  it  is  the  hope 
of  the  authors  that  this  stimulus  may  not  be  lacking  when 
the  volume  is» studied. 

New  Haven,  Connecticut 


CONTENTS 


CHAPTER  I 
FORMULAS  FOR  REFERENCE 

SECTION  PAGE 

1 .  Formulas  and  theorems  from  geometry,  algebra,  and  trigonometry  1 

2.  Three-place  table  of  common  logarithms  of  numbers       ....  4 

3.  Logarithms  of  trigonometric  functions 5 

4.  Natural  values  of  trigonometric  functions 5 

5.  Special  angles.    Natural  values 6 

6.  Rules  for  signs 7 

7.  Greek  alphabet 7 

CHAPTER   II 
CARTESIAN  COORDINATES 

8.  Directed  line 8 

9.  Cartesian  coordinates 9 

10.  Rectangular  coordinates 10 

11.  Lengths 13 

12.  Inclination  and  slope 16 

13.  Areas 21 

CHAPTER  III 
CURVE  AND  EQUATION 

14.  Locus  of  a  point  satisfying  a  given  condition 28 

15.  Equation  of  the  locus  of  a  point  satisfying  a  given  condition  .    .  28 

16.  First  fundamental  problem 30 

17.  Locus  of  an  equation 35. 

18.  Second  fundamental  problem 36 

19.  Third  fundamental  problem.    Discussion  of  an  equation      ...  41 

20.  Symmetry 45 

21.  Further  discussion 46 

22.  Directions  for  discussing  an  eqviation 47 

23.  Points  of  intersection 50 

vii 


viii  ELEMENTARY  ANALYSIS 

CHAPTER   IV 
STRAIGHT  LINE  AND  CIRCLE 

SECTION  PAGE 

24.  The  degree  of  the  equation  of  any  straight  line '.    .  54 

25.  Locus  of  any  equation  of  the  first  degree 55 

26.  Plotting  straight  lines 56 

27.  Point-slope  equation 58 

28.  Two-point  equation 59 

29.  The  angle  which  a  line  makes  with  a  second  line 63 

30.  Equation  of  the  circle 68 

31.  Circles  determined  by  three  conditions 70 

Locus  problems 74 

CHAPTER  V 
CURVE  PLOTTING 

32.  Asymptotes 77 

33.  Natural  logarithms ' 79 

Exponential  and  logarithmic  curves 80 

Compound  interest  curve 82 

34.  Sine  curves 83 

35.  Addition  of  ordinates 88 

CHAPTER   VI 
FUNCTIONS  AND  GRAPHS 

36.  Functions 91 

37.  Notation  of  functions 100 

CHAPTER   VII 
DIFFERENTIATION 

38.  Tangent  at  a  point  on  the  graph 102 

39.  Differentiation 103 

40.  Derivative  of  a  function 105 

CHAPTER  VIII 
FORMULAS  FOR  DIFFERENTIATION 

41.  Theorems  on  limits Ill 

42.  Fundamental  formulas 113 


CONTENTS  ix 

SECTION  PAGE 

43.  Differentiation  of  a  constant 114 

44.  Differentiation  of  a  variable  with  respect  to  itself 114 

45.  Differentiation  of  a  sum 115 

46.  Differentiation  of  the  product  of  a  constant  and  a  function  .    .  115 

47.  Differentiation  of  the  product  of  two  functions 116 

48.  Differentiation  of  a  power  of  a  function 117 

49.  Differentiation  of  a  quotient 117 

50.  Differentiation  of  a  function  of  a  function  . 123 

51.  Differentiation  of  a  logarithm 124 

52.  Differentiation  of  an  exponential  function 127 

53.  Proof  of  the  power  rule 129 

54.  Differentiation  of  sin  u 132 

55.  Differentiation  of  cos  v 135 

56.  Differentiation  of  tan  .u : 135 

57.  Proofs  of  XIII-XV 136 

58.  Inverse  circular  functions 139 

59.  Differentiation  of  sin-iv  and  tan- 1  u 140 

60.  Implicit  functions 143 


CHAPTER   IX 
SLOPE,  TANGENT,  AND  NORMAL 

61.  Tangent  and  normal 146 

CHAPTER  X 
MAXIMA  AND  MINIMA 

62.  Rule  for  the  determination  of  maximum  and  minimum  values  .  151 

63.  Derivatives  of  higher  orders 156 

64.  Geometrical  significance  of  the  second  derivative 158 

65.  Second  test  for  maxima  and  minima 161 

m.  Points  of  inflection , 162 

CHAPTER   XI 

RATES 

67.  Velocity  and  acceleration     .   ^. 167 

Time  rates •    •    168 


ELEMENTAEY  AI^ALYSIS 


CHAPTER  XII 
DIFFERENTIALS 

SECTION  PAGE 

68.  Definition  of  differential 175 

69.  Formulas  for  finding  differentials 176 

70.  Infinitesimals 179 

CHAPTER  XIII 

INTEGRATION.    RULES  FOR  INTEGRATING  STANDARD 
ELEMENTARY  FORMS 

71.  Integration 180 

72.  Constant  of  integration.    Indefinite  integral 182 

73.  Rules  for  integrating  standard  elementary  forms 183 

74.  Trigonometric  and  other  substitutions 194 

CHAPTER  XIV 
CONSTANT  OF  INTEGRATION 

75.  Determination  of  the  constant  of  integration  by  means  of  in- 

itial conditions 196 

Compound  interest  law 197 

CHAPTER   XV 
THE  DEFINITE  INTEGRAL 

76.  Differential  of  an  area 201 

77.  The  definite  integral 202 

78.  Geometric  representation  of  an  integral      .    . 204 

79.  Calculation  of  a  definite  integral 204 

CHAPTER   XVI 
INTEGRATION  A  PROCESS  OF  SUMMATION 

80.  Introduction 207 

81.  The  fundamental  theorem  of  the  integral  calculus 207 

82.  Areas  of  plane  curves 211 

83.  Volumes  of  solids  of  revolution 216 

84.  Miscellaneous  applications  of  the  integral  calculus 219 


ELEMENTARY  Al^ALYSIS 

CHAPTER   I 

FORMULAS  FOR   REFERENCE 

I.  Occasion-55^ill  arise  in  later  chapters  to  make  use  of  the 
following  formulas  and  theorems  proved  in  geometry,  algebra, 
and  trigonometry. 

1.  Circumference  of  circle  =  2  7rr.*       3.   Volume  of  prism  =  Ba. 

2.  Area  of  circle  =  irr^.  4.   Volume  of  pyramid  =  J  Ba. 

5.  Volume  of  right  circular  cylinder  =  wr'^a- 

6.  Lateral  surface  of  right  circular  cylinder  =  2  irra. 

7.  Total  surface  of  right  circular  cylinder  =  2  7r7'(r  4-  a). 

8.  Volume  of  right  circular  cone  =  ]  irr^a. 

9.  Lateral  surface  of  right  circular  cone  =7rrs. 

10.   Total  surface  of  right  circular  cone  =  7rr(r  -f-  s) . 

II.  Volume  of  sphere  :=  |  irr^.  12.    Surface  of  sphere  =  4  Trr^. 

13.  In  a  geometrical  series, 

r  —  1  r  —  1 

a  =  first  term,  r  =  common  ratio,  I  =  nth  term,  s  =:  sum  of  n  terms. 

14.  loga?)  =  loga +log&.  *     16.   loga^  =  wloga. 


-  \o^  a. 


*In  formulas  1-12,  r  denotes  radius,  a  altitude,  B  area  of  base,  and  .9  slant 
height. 

1 


15. 

I0g^  =  l0grt- 

b 

-  log  b. 

17. 

log  Va  =-\oga. 
n 

18. 

log  1=0. 

19. 

logaa  =  1. 

20.   logi 

V.-'«--7>rv^^  V       \*i'0-v-«iafc-*. 


ELEMENTARY    ANALYSIS 

Functicns  of  an  aixjU  In  a  ri^p  triangle.    In  any  right  triangle 
of  whose  acute  angles  is  A^  the  functions  of  A  are  defined  as  follows ; 

_  opposite  side 


21. 


sin  A  —  -— , 

hypotenuse 

cos^  =  ^dj^cent  side 
hypotenuse 

^^  ^  ^  opposite^ide 
adjacent  side ' 


CSC  A  =  ^^yPQtenuse  ^ 
opposite  side' 

sec  A  =  ^^yPQ^enuse  ^ 
adjacent  side ' 

.    A  _  adjacent  side 
opposite  side 


From  the  above  the  theorem  is  easily  derived : 

22.  In  a  right  triangle  a  side  is  equal  to  the  prod- 
uct of  the  hypotenuse  and  the  sine  of  the  angle  op- 
posite to  that  side,  or  of  the  hypotenuse  and  the 
cosine  of  the  angle  adjacent  to  that  side. 

Angles  in   general.    In  Trigonometry  an  angle 

XOA  is  considered  as 

generated  by  the  line  ^ 

OA  rotating  from  an 
initial  position  OX.  The  angle  is  positive 
when  OA  rotates  from  OX  counter-clock- 
loise^  and  negative  when  the  direction  of 
rotation  of  OA  is  clockivise. 

The  fixed  line  OX  is  called  the  initial 
line,  the  line  OA  the  terminal  line. 

Measurement. of  angles.  There  are  two  important  methods  of  measur- 
ing angular  magnitude,  that  is,  there  are  two  unit  angles. 

Degree  measure.  The  unit  angle  is  ^i^  of  a  complete  revolution,  and  is 
called  a  degree. 

Circular  measure.  The  unit  angle  is  an  angle  whose  subtending  arc  i^ 
equal  to  the  radius  of  that  arc,  and  is  called  a  radian. 

The  fundamental  relation  between  the  unit  angles  is  given  by  the 
equation 

23.  180  degrees  =  ir  radians  (ir^:  3.14159  •••)• 
Or  also,  by  solving  this, 

24.  1  degree  =  -^  =  .0174  •  •  •  radians. 

^  180 


owa* 


25. 


1  radian : 


180 


:  57.29  ...  degrees. 


FORMULAS   FOR   REFERENCE  3 

These  equations  enable  us  to  change  from  one  measurement  to  another. 
In  the  higher  mathematics  circular  measure  is  always  used,  and  will  be 
adopted  in  this  book. 

The  generating  line  is  conceived  of  as  rotating  around  0  through  as 
many  revolutions  as  we  choose.     Hence  the  important  result : 

Any  real  number  is  the  circular  measure  of  some  angle,  and  conversely, 
any  angle  is  measured  by  a  real  number. 

26.  cotx=: ;secic=: ;cscic= 

tan  X  cos  x  sin  x 

27.  tana;  =  ?i^;  cotx  =  55?^. 

cos  X  sin  X 

28.  sin2  X  +  cos2  x  =  1  ;  1  +  tan^x  =  sec%  ;  1  +  cot^ic  =  csc^  x. 

29.  sin  (—  x)  =  —  sin  x]  esc  (—  x)  =  —  esc  x  ; 
cos  ( —  x)  =  cos  a:; ;  sec  (—  a:)  =  sec  x  ; 
tan  (  —  x)  =  —  tan  x ;  cot  (  —  x)  =  —  cot  x. 

30.  sin  (tt  —  x)  =  sin  x  ;  sin  (tt  +  x)  =  —  sin  x  ; 
cos  (tt  —  x)  =  —  cos  X  ;  cos  (jr  -{-x)  =  —  cos  x ; 
tan  (tt  —  x)  =  —  tan  x  ;  tan  (tt  +  x)  =      tan  x. 

31.  sin  r^-— X  j  =  cosx;  sin(  ^+ X )  =      cosx; 

^  —  X  J  =:  sin  X ;  cos  (^+xj  =  —  sinx; 
t;i !  1  (  ^  —  X  1  =  cot  X  ;  tan  f  '^  +  x  )  =  —  cot  x. 

1.2     y         '     V2     y 

2  TT  —  x)  =  sin(—  x)  =  —  sin  x,  etc. 
•  >:  +  ?/)  =  sin  x  cos  ?/  +  cos  x  sin  ?/. 

X  —  y)—  sin  x  cos  ?/  —  cos  x  sin  ,'/. 
X  +  ?/)  =  cos  X  cos  ?/  —  sin  x  sin  i' , 
X—  y)=  cos  X  cos  y  +  sin  x  sin  y. 

'  ,       .  +  y)=:   t^^^  +  tany  ^      38.  tan  (x  -  y)  ^   tan  x  -  tan  y_ 

1  -  tan  X  tan  ?/  1  +  tan  x  tan  ?/ 

39.    i  ■ !; :  c  =:  2  sin  x  cos x  ;  cos  2  x  =  cos'^  x  —  sin^  x  ;  tan  2  x  =  _  2  tan  x 


ELEMENTARY   ANALYSIS 


40, 


'      X  ,       /l  —  cos  X  X         ,       ll  -\-  COS  X      .         X  ,       /l  —  COS  X 


+  COSX 


2  >(        2  2  >(        2 

41.  sin2  X  =  I  ~  I  COS  2  x  ;  cos'^  x  =  ^  +  I  cos  2  x. 

42.  sin  A  -  sin  ^  =  2  cos  ^  (^  +  ^)  sin  i  (A  -  B), 

43.  cos  ^  -  cos  jB  =—  2  sin  ^  (A -{-  B)  sin  |  (^  -  ^)- 

44.  Theorem.  Law  of  cosines.  In  any  triangle  the  square  of  a  side 
equals  the  sum  of  the  squares  of  the  two  other  sides  diminished  by  twice 
the  product  of  those  sides  by  the  cosine  of  their  included  angle ; 

that  is,  a2  =  &2  ^_  c2  _  2  &c  cos  A. 

45.  Theorem.     Area  of  a  triangle.     The  area  of  any  triangle  equals 
one  half  the  product  of  two  sides  by  the  sine  of  their  included  angle ; 
that  is,  area  =  |  a6  sin  C  =  J  6c  sin  A  =  ^  ca  sin  B, 


2.  Three-place  table  of  common  logarithms  of  numbers. 


N 

0 

1 

2 

3 

4 

5 

6 

7 

« 

9 

1 

000 

041 

079 

114 

146 

176 

204 

230 

255 

279 

2 

301 

322 

342 

362 

380 

398 

415 

431 

447 

462 

3 

477 

491 

505 

518 

532 

544 

556 

568 

58" 

•  -' 

4 

602 

613 

623 

634 

643 

653 

663 

672 

6^ 

5 

099 

708 

716 

724 

732 

740 

748 

756 

703 

771 

6 

778 

785 

792 

799 

806 

813 

820 

826 

832 

8ro 

7 

845 

851 

857 

863 

869 

875 

881 

886 

S-- 

8 

903 

908 

914 

919 

924 

929 

934 

939 

9-t  + 

9 
10 

11 

954 

959 

964 

968 

973 

978 

982 

987 

037 

000 

004 

009 

013 

017 

021 

025 

029 

041 

045 

049 

053 

057 

061 

064 

068 

072 

076 

12 

079 

083 

08^5 

093 

097 

100 

104 

107 

ill 

13 

114 

117 

121 

127 

130 

134 

137 

140 

143 

14 

146 

149 

152 

155 

158 

161 

164 

167 

170 

173 

15 

176 

179 

182 

185 

188 

190 

193 

196 

190 

201 

16 

204 

207 

210 

212 

215 

218 

220 

223 

22o 

228 

17 

230 

233 

236 

238 

241 

243 

246 

248 

250 

253 

18 

255 

258 

260 

262 

265 

267 

270 

272 

274 

276 

19 

279 

281 

283 

286 

288 

290 

292 

294 

297 

299 

FORMULAS   FOR  REFERENCE 


3.   Logarithms  of  trigonometric  functions. 


Angle  in 
Radians 

Angle  in 
Degrees 

log  sin 

log  cos 

log  tan 

log  cot 

.000 

0° 

0.000 



90° 

1.571 

.017 

1° 

8.242 

9.999 

8.242 

1.758, 

89° 

1.553 

.035 

2° 

8.543 

9.999 

8.543 

1.457 

88° 

1.536 

.052 

3^^ 

8.719 

9.999 

8.719 

1.281 

87° 

1.518 

.070 

40 

8.844 

9.999 

8.845 

1.155 

86° 

1.501 

.087 

5° 

8.940 

9.998 

8.942 

1.058 

85° 

1.484 

.174 

10" 

9.240 

9.993 

9.246 

0.754 

80° 

1.396 

.262 

15° 

9.413 

9.985 

9.428 

0.572 

75° 

1.309 

.349 

20° 

9.534 

9.973 

9.561 

0.439 

70° 

1.222 

.436 

25° 

9.026 

9.957 

9.609 

0.331 

65° 

1.134 

.524 

30° 

9.699 

9.938 

9.761 

0.239 

60° 

1.047 

.611 

35° 

9.759 

9.913 

9.845 

0.165 

55° 

0.960 

.698 

40° 

9.808 

9.884 

9.924 

0.080 

50° 

0.873 

.785 

45° 

9.850 

9.850 

0.000 

0.000 

45° 

0.785 

log  cos 

log  sin 

log  cot 

log  tan 

Angle  in 
Degrees 

Angle  in 
Radians 

4.   Natural  values  of  trigonometric  functions. 

Angle  in 
Radians 

Angle  in 
Degrees 

sin 

COS 

tan 

cot 

.000 

0° 

.000 

1.000 

.000 

QO 

90° 

1.571 

.017 

1° 

.017 

.999 

.017 

57.29 

89° 

1.553 

.035 

2° 

.035 

.999 

.035 

28.64 

88° 

1.536 

.052 

30 

.052 

.999 

.052 

19.08 

sr 

1.518 

.070 

40 

.070 

.998 

.070 

14.30 

86° 

1.501 

.087 

5° 

.087 

.996 

.088 

11.43 

85° 

1.484 

.174 

10° 

.174 

.985 

.176 

5.67 

80° 

1.396 

.262 

15° 

.259 

.966 

.268 

3.73 

75° 

1.309 

.349 

20° 

.342 

.940 

.364 

2.75 

70° 

1.222 

.436 

25° 

.423 

.906 

.466 

2.14 

65° 

1.134 

.524 

30° 

.500 

.866 

.577 

1.73 

60° 

1.047 

.611 

35° 

.574 

.819 

.700 

1.43 

55° 

.960 

.698 

40° 

.643 

.766 

.839 

1.19 

50° 

.873 

.785 

45° 

.707 

.707 

1.000 

1.00 

45° 

.785 

cos 

sin 

cot 

tan 

Angle  in 
Degrees 

Angle  in 
Radians 

6  ELEMENTARY   ANALYSIS 

5.   Special  angles.     Natural  values. 


Angle  in 
Radians 

Angle  in 
Degrees 

sin 

COS 

tan 

cot 

sec 

CSC 

0 

0° 

0 

1 

0 

GO 

1 

GO 

IT 

2 

90^ 

1 

0 

GO 

0 

GO 

1 

IT 

180^ 

0 

-  1 

0 

00 

-   1 

GO 

Sir 

2 

270^^ 

-1 

0 

GO 

0 

00 

-  1 

27r 

360° 

0 

1 

0 

00 

1 

00 

Angle  in 
Radians 

Angle  in 
Degrees 

sin 

cos 

tan 

cot 

sec 

CSC 

0 

0° 

0 

1 

0 

- ;    GO 

1 

GO 

TT 

6 

30^=^ 

1 
2 

2 

3 

'  V.^.,  ■- 

3 

2 

IT 

4 

45'^ 

V2 

2 

V2    - 

2 

1 

1 

V2 

V2 

TT 
3 

60° 

V3 
2 

1 

2 

V3 

V3 
3 

2 

2V3 
3 

TT 

2 

90° 

1 

0 

00 

0 

GO 

1 

The  student  sliould  understand  that  the  numerical  values  of 
the  functions  given  in  the  second  of  the  above  tables  are  to  be 
used  when  formal  exactness  only  is  required ;  that  is,  when  it 
is  sufficient  to  leave  the  results  in  radical  form.  Explicit 
numerical  values  are  given  in  Art.  4.  The  same  remarks 
apply  to  the  values  of  the  angle  in  radians. 


FORMULAS   FOR   REFERENCE 
6.   Rules  for  signs.         ^  __ 


-V         - 


Quadrant 

sin 

cos 

tan 

cot 

sec 

CSC 

First 

+ 

+ 

+ 

+ 

+ 

+ 

Second  .... 

+ 

- 

- 

- 

+ 

Third     .... 

- 

- 

+ 

+ 

- 

- 

Fourth  .... 

- 

+ 

- 

- 

+ 

- 

7.   Greek  alphabet. 


ETTERS 

Names 

Letters 

Names 

Letters 

Names 

Aa 

Alpha 

I    I 

Iota 

PP 

Rho 

B^ 

Beta 

K/c 

Kappa 

S  (7   S 

Sigma 

Ft 

Gamma 

A  \ 

Lambda 

T   T 

Tau 

A5 

Delta 

M/x 

Mu 

Tu 

Upsilon 

E  e 

Epsilon 

N  V 

Nil 

*  0 

Phi 

zr 

Zeta 

s^ 

Xi 

Xx 

Chi 

H^ 

Eta 

Oo 

0  micron 

<v^ 

Psi 

e6> 

Theta 

Htt 

Pi 

12  w 

Omega 

CHAPTER   II 

CARTESIAN   COORDINATES 

8.  Directed  line.  Let  X'X  be  an  indefinite  straight  line, 
and  let  a  point  0,  which  we  shall  call  the  origin,  be  chosen 
upon  it.  Let  a  unit  of  length  be  adopted,  and  assume  that 
lengths  measured  from  0  to  the  right  are  x>ositive,  and  to  the 
left  negative. 

-5-4-3-2-1    0-hl-+2+3-l-4+5  unit 


X'  O        ^  X      1 

Then  any  real  number,  if  taken  as  the  measure  of  the  length 
of  a  line  OP,  will  determine  a  point  P  on  the  line.  Con- 
versely, to  each  point  P  on  the  line  will  correspond  a  real 
number ;  namely,  the  measure  of  the  length  OP,  with  a  posi- 
tive or  negative  sign  according  as  P  is  to  the  right  or  left  of 
the  origin. 

The  direction  established  upon  X^X  by  passing  from  the 
origin  to  the  points  corresponding  to  the  positive  numbers  is 
called  the  positive  direction  on  the  line.     A  directed  line  is  a 


~B  2  0  X        "b^ 

straight  line  upon  which  an  origin,  a  unit  of  length,  and  a 
positive  direction  have  been  assumed. 

An  arrowhead  is  usually  placed  upon  a  directed  line  to  indi- 
cate the  positive  direction. 

If  A  and  B  are  any  two  points  of  a  directed  line  such  that 

OA  =  a,    OB  =  b, 

then  the  length  of  the  segment  AB  is  always  given  by  b  —  a-, 
that  is,  the  length  of  AB  is  the  difference  of  the  numbers  cor- 

8 


CARTESIAN   COORDINATES  9 

responding  to  B  and  A.  This  statement  is  evidently  equiva- 
lent to  the  following  definition : 

For  all  positions  of  two  p)oints  A  and  B  on  a  directed  line,  the 
length  AB  is  given  by 

(1)  AB^OB-OA, 

where  O  is  the  origin, 

(I)  Cll)  an)  (IV) 

0  +3  -h6       -4  0+3        -3      0         H-5       -6      -2    0 

O     4.  M        B       0      A        A     0  B        B      A  0 

Illustrations. 

In  Fig.  ^  ^- 

I.  ^li?=  01^-0^^=0-3= +3;  ^^=0^-05=3-6  =  -3; 
II.  AB=0B-0A=-4.-?,=  -l',  75^=0^-05=3-(-4)  =  +7; 

III.  ^5=0i>^-0.1=+5-(-3)  =  +8;  ^^l=r0.4-OB=-3-5  =  -8; 

IV.  AB=OB-OA=-Q-{-2)  =  -4:-  BA^0A-0B=-2-{-&)^+^, 

The  following  properties  of  lengths  on  a  directed  line  are 
obvious : 

(2)  AB  =-BA. 

(3)  AB  is  positive  if  the  direction  from  A  to  B  agrees  with 
the  positive  direction  on  the  line,  and  negative  if  in  the  con- 
trary direction. 

The  phrase  "  distance  between  two  points  "  should  not  be  used  if  these 
points  lie  upon  a  directed  line.  Instead,  we  speak  of  the  length  AB^  re- 
membering that  the  lengths  AB  and  BA  are  not  equal,  but  that  AB  =  — 
BA. 

9.   Cartesian^  coordinates.      Let  X'X  and   Y'Y  be  two 

directed  lines  intersecting  at  0,  and  let  P  be  any  point  in 
their  plane.  Draw  lines  through  P  parallel  to  X'X  and  Y'  Y 
respectively.     Then,  if 

OM=a,    ON=h, 

*  So  called  after  Eene  Descartes,  ISOB-IOHO,  who  first  introduced  the  idea 
of  coordinates  into  the  study  of  Geometry. 


10 


ELEMENTARY  ANALYSIS 


the  numbers  a,  b  are  called  the  Cartesian  coordinates  of  P,  a 
the  abscissa  and  b  the  ordinate.     The  directed  lines  X'X  and 

F'  Y  are  called  the  axes 
of  coordinates,  X'X  the 
axis  of  abscissas,  Y'Y the 
axis  of  ordinates,  and 
their  intersection  0  the 
origin. 

The  coordinates  a,  b 
of  P  are  written  (a,  b), 
and  the  symbol  P(a,  b) 
is  to  be  read:  "The 
point  P,  whose  coordi- 
nates are  a  and  b^ 
Any  point  P  in  the  plane  determines  two  numbers,  the 
coordinates  of  P.  Conversely,  given  two  real  numbers  a'  and 
5',  then  a  point  P'  in  the  plane  may  always  be  constructed 
whose  coordinates  are  (a',  V).  For  lay  off  OM^  =  a',  0N^=  b', 
and  draw  lines  parallel  to  the  axes  through  JP  and  N\  These 
lines  intersect  at  P\a\  b').     Hence 

Every  point  determines  a  pair  of  real  numbers^  and  conversely, 
a  pair  of  real  numbers  determines  a  point. 

The  imaginary  numbers  of  Algebra  have  no  place  in  this 
representation,  and  for  this  reason  elementary  Analytic  Geome- 
try is  concerned  only  with  the  real  numbers  of  Algebra. 

10.  Rectangular  coordinates.  A  rectangular  system  of 
coordinates  is  determined  when  the  axes  X^X  and  F'  Y  are 
perpendicular  to  each  other.  This  is  the  usual  case,  and  will 
be  assumed  unless  otherwise  stated. 

The  work  of  plotting  points  in  a  rectangular  system  is  much 
*  simplified  by  the  use  of  coordinate  ov  plotting  paper,  constructed 
by  ruling  off  the  plane  into  equal  squares,  the  sides  being  parallel 
to  the  axes. 

In  the  figure  several  points  are  plotted,  the  unit  of  length 


CARTESIAN   COORDINATES 


11 


> 

Y 

(6, 

7) 

{-4 

6) 

X' 

0 

qo, 

0) 

X 

{-0 

-4) 

{0, 

-4) 

I  1 

being  assumed  equal  to  one  division  on  each  axis.     The  method 
is  simply  this: 

Count  off  from  0  along  X'X  a  number  of  divisions  equal  to 
the  given  abscissa,  and  then  from  the  point  so  determined  a 
number  of  divisions  equal  to  the  given  ordinate,  observing  the 

Rule  for  signs : 

Abscissas  are  positive  or  negative  according  as  they  are  laid  off 
to  the  right  or  left  of  the  origin,  Ordinates  are  positive  or  nega- 
tive according  as  they  are  laid  off  above  or 
beloiv  the  axis  of  x. 

Eectangular  axes  divide  the  plane  into 
four  portions  called  quadrants;  these  are 
numbered  as  in  the  figure,  in  vs^hich  the 
proper  signs  of  the  coordinates  are  also 
indicated. 


rt 


Second 


First 
(f  v-f) 


X 


Fourth 


X'  0 

Thxrd 
(-,-) 

Y' 

As  distinguished  from  rectangular  coordinates,  the  term 
oblique  coordinates  is  employed  when  the  axes  are  not  perpen- 
dicular, as  in  the  figure,  p.  10.  The  rule  of  signs  given  above 
applies  to  this  case  also. 

In  the  following  problems  assume  rectangular  coordinates. 


12  ELEMENTARY   ANALYSIS 

PROBLEMS 

1.  Plot  accurately  the  points  (3,  2),  (3,  -  2),  (-4,  3),  (6,  0), 
(-5,0),  (0,4). 

2.  What  are  the  coordinates  of  the  origin  ?  Ans.  (0,  0). 

3.  In  what  quadrants  do  the  following  points  lie  if  a  and  b 
are  positive  numbers:  (—a,  b)?  (—a,  —b)?  {b;  —  a)?  (a\  b)? 

4.  To  what  quadrants  is  a  point  limited  if  its  abscissa  is 
positive  ?  negative  ?  its  ordinate  is  positive  ?  negative  ? 

5.  Plot  the  triangle  whose  vertices  are  (2,  —1),  (—2,  5), 
(-8,-4). 

6.  Plot  the  triangle  whose  vertices  are  ( —  2,  0),  (5^3—2,  5), 
(-2,10). 

7.  Plot  the  quadrilateral  whose  vertices  are  (0,  —  2),  (4, 2), 
(0,6),  (-4,2). 

8.  If  a  point  moves  parallel  to  the  axis  of  x,  which  of  its 
coordinates  remains  constant?  if  parallel  to  the  axis  of  y? 

9.  Can  a  point  move  when  its  abscissa  is  zero  ?  Where  ? 
Can  it  move  when  its  ordinate  is  zero  ?  Where  ?  Can  it  move 
if  both  abscissa  and  ordinate  are  zero  ?     Where  will  it  be  ?  p 

10.  Where  may  a  point  be  found  if  its  abscissa  is  2  ?  if  its 
ordinate  is  —  3? 

r.     11.    Where  do  all  those  points  lie  whose  abscissas  and  ordi- 
iiates  are  equal  ? 

12.  Two  sides  of  a  rectangle  of  lengths  a  and  b  coincide  with 
the  axes  of  x  and  y  respectively.  What  are  the  coordinate^ 
of  the  vertices  of  the  rectangle  if  it  lies  in  the  first  quadrant  ?  in 
the  second  quadrant  ?  in  the  third  quadrant  ?  in  the  fourth 
quadrant  ? 

13.  Construct  the  quadrilateral  whose  vertices  are  (—3,  6), 
(_3^  0),  (3,  0),  (3,  6).     What  kind  of  a  quadrilateral  is  it? 

14.  Show  that  {x,  y)  and  {x,  '~y)  are  symmetrical  with  re- 
spect to  X'X;  (Xj  y)  and  {— x,  y)  with  respect  to  F'F;  and 
{x,  y)  and  {  —  x^  ~  V)  with  respect  to  the  origin. 


CARTESIAN   COORDINATES 


13 


15.  A  line  joining  two  points  is  bisected  at  the  origin.  If 
the  coordinates  of  one  end  are  {a,  —  b),  what  will  be  the  coor- 
dinates of  the  other  end  ? 

16.  Consider  the  bisectors  of  the  angles  between  the  coordi- 
nate axes.  What  is  the  relation  between  the  abscissa  and 
ordinate  of  any  point  of  the  bisector  in  the  first  and  third 
quadrants  ?  second  and  fourth  quadrants  ? 

17.  A  square  whose  side  is  2  a  has  its  center  at  the  origin. 
What  will  be  the  coordinates  of  its  vertices  if  the  sides  are 
parallel  to  the  axes  ?  if  the  diagonals  coincide  with  the  axes  ? 

Arts,    (a,  a),  (c(,  —  a)^  (—  a,  —  a),  (—  a,  a) ; 

(a V2,  0),  (-  a V2,  0),  (0,  aV2),  (0,  -  aV2). 

18.  An  equilateral  triangle  whose  side  is  a  has  its  base  on 
the  axis  of  x  and  the  opposite  vertex  above  X'X.  What  are 
the  vertices  of  the  triangle  if  the  center  of  the  base  is  at  the 
origin  ?  if  the  lower  left-hand  vertex  is  at  the  origin  ? 


11.    Lengths.     Consider  any  two  given  points 

Pli^iy  Vl),  ^2(^2,  2/2). 

'      Then  in  the  figure  03f^  =  x\,  03L  = 

We  may  now  easily  prove  the  im- 
portant 

Theorem.  The  length  I  of  the  line  j^t-q 
joining  ttvo  points  Pi{x^,  2/i)?  ^2(^^27  2/2)  ^^  ^J 
given  by  the  formula 

(I)  l  =  V{xi^'jc2y  +  {yi-y2y- 

Proof     Draw  lines  through  P^  and  Po  parallel  to  the  axes  to 
form  the  right  triangle  P^iSPu 


Mnt^-^^. 


14 


ELEMENTARY   ANALYSIS 


Then 


and  hence 


P1P2  =  '^jSP2^  +  P^S' 


Q.E.D. 


(-5, 


The  same  method  is  used  in  deriving  (I)  for  any  positions 
of  Pi  and  P2 ;  namely,  we  construct  a  right  triangle  by  draw- 
ing lines  parallel  to  the  axes  through  P^  and  Pg-  The  hori- 
zontal side  of  this  triangle  is  equal  to  the  difference  of  the 
abscissas  of  P^  and  P2,  while  the  vertical  side  is  equal  to  the 
difference  of  the  ordinates.  The  required  length  is  then 
the  square  root  of  the  sum  of  the  squares  of  these  sides,  which 
gives  (I).  A  number  of  different  figures  should  be  drawn  to 
make  the  method  clear. 

EXAMPLE 

Find  the  length  of  the  line  join- 
ing the  points  (1,  3)  and(—  5,  5). 

Solution.      Call    (1,   3)    Pj,   and 

(-5,5)P,. 

Then 
Xi  =  1,  ?/i  =  3,  and  x2  =  —  5,y2  =  5i 
and  substituting  in  (I),  we  have 
I  =  V(l  +  5/+(3-5)2  =  V40  =  2VIO. 

It  should  be  noticed  that  we  are  simply  finding  the  hypote- 
nuse of  a  right  triangle  whose  sides  are  6  and  2. 

Remark,  The  fact  that  formula  (I)  is  true  for  all  positions 
of  the  points  Pi  and  P2  is  of  fundamental  importance.  The 
application  of  this  formula  to  any  given  problem  is  therefore 
simply  a  matter  of  direct  substitution.  In  deriving  such  gen- 
eral formulas,  it  is  most  convenient  to  draw  the  figure  so  that 
the  points  lie  in  the  first  quadrant,  or,  in  general,  so  that  all 
the  quantities  assumed  as  known  shall  he  positive. 


(H 


X 


CARTESIAN   C0ORMN  ATES  U        /  15 

PROBLEMS 

1.  Find  the  projections  on  the  axes  and  the  length  of  the 
lines  joining  the  following  points. 

(a)  (-4,  -4)  and  (1,3). 

Ans.   Projections  5,  7 ;  length  =  V74. 

(b)  (- V2,  V3)  and  (V3,  V2). 

Ans.   Projections  V3  +  V2,  V2  -  V3 ;  length  =  VlO. 

(0   (0,  0)  and  g,  -^y 

Ans.    Projections  -,  -V3;  length  =  a. 

(d)  (a  -i-hj  G  +  a)  and  (c  +  ay  6  +  c). 

Ans.    Projections  c  —  b,  b  —  a-,  length  =  V(6  —  c)^+  (a  —  b^). 

2.  Find  the  projections  of  the  sides  of  the  following  tri- 
angles upon  the  axes : 

(^)  (0,  6),  (1,  2),  (3,  -  5). 

(b)  (1,0),(~1,  -5),(-l,  -8). 

(c)  (a,  b),  (b,  c),  (c,  (^). 

(d)  (a,  -  b),  (b,  -  c),  (c,  -  d). 

(e)  (0,7y),  (-^,  -y)^^-x,0). 

3.  Find    the    lengths    of    the    sides    of    the    triangles    in 
problem  2. 

4.  Work  out  formula  (I)  :   (a)  if  0^1  =  0*2;  (6)  if?/i=2/2. 

5.  Find  the  lengths  of  the  sides  of  the  triangle  whose  ver- 
tices are  (4,  3),  (2,  -2),  (-3,  5). 

6.  Show  that  the  points  (1,  4),  (4,  1),  (5,  5)  are  the  vertices 
of  an  isosceles  triangle. 

7.  Show  that  the  points  (2,  2),  (-2,  -2),  (2V3,  -2V3) 
are  the  vertices  of  an  equilateral  triangle. 

8'.    Show  that  (3,  0),  (6,  4),  (—1,  3)  are  the  vertices  of  a 
right  triangle.     What  is  its  area  ? 


16  ELEMENTARY  ANALYSIS 

9.  Prove  that  (-  4,  -  2),  (2,  0),  (8,  6),  (2,  4)  are  the  ver- 
tices of  a  parallelogram.  Also  find  the  lengths  of  the 
diagonals. 

10.  Show  that  (11,  2),  (6,  -10),  (-6,  -5),  (-1,  7)  are 
the  vertices  of  a  square.     Find  its  area. 

11.  Show  that  the  points  (1,  3),  (2,  V6),  (2,  —  V6)  are  equi- 
distant from  the  origin ;  that  is,  show  that  they  lie  on  a  circle 
with  its  center  at  the  origin  and  its  radius  VlO. 

12.  Show  that  the  diagonals  of  any  rectangle  are  equal. 

13.  Find  the  perimeter  of  the  triangle  whose  vertices  are 
(a,  6),  (-a,  b),  (-<i,  -b). 

14.  Find  the  perimeter  of  the  polygon  formed  by  joining 
the  following  points  two  by  two  in  order : 

(6,4),  (4,  -3),  (0,  -1),  (-6,  -4),  (-2,1). 

15.  One  end  of  a  line  whose  length  is  13  is  the  point 
(—4,8);  the  ordinate  of  the  other  end  is  3.  What  is  its 
abscissa  ?  Aiis.   8  or  —  16. 

16.  What  equation  must  the  coordinates  of  the  point  (.^',  y) 
satisfy  if  its  distance  from  the  point  (7,  —  2)  is  equal  to  11  ? 

17.  What  equation  expresses  algebraically  the  fact  that  the 
point  (Xf  y)  is  equidistant  from  the  points  (2,  3)  and  (4,  5)  ? 

12.  Inclination  and  slope.  The  angle  between  two  inter- 
secting directed  lines  is  defined  to  be  the  angle 
made  by  their  positive  directions.  In  the 
figures  the  angle  between  the  directed  lines  is 
the  angle  marked  6. 

If  the  directed  lines  are  parallel,  then  the 
angle  between  them  is  zero  or  ir  according  as 
the  positive  directions  agree  or  do  not  agree. 

Evidently  the  angle  between  two  directed 
lines  may  have  any  value  from  0  to  tt  in- 
clusive.    Reversinjjr  the  direction  of  either  di- 


CARTESIAN   COORDINATES 


17 


rected  line  changes  0  to  the  supplement  tt  - 
tions  are  reversed,  the  angle  is  unchanged. 


0.     If  both  direc- 


^  =  0 


When  it  is  desired  to  assign  a  positive  direction  to  a  line 
intersecting  X'X,  we  shall  always  assume  the  upimrd  direction 
as  positive. 

The  inclination  of  a  line  is  the  angle  between  the  axis  of  x 
and  the   line  when    the   latter   is 
given  the  upward  direction. 

The  sloj^e  of  a  line  is  the  tan- 
gent of  its  inclination. 

The  inclination  of  a  line  will  be 
denoted  by  a,  a^,  a^,  a\  etc.;  its 
slope  by  m,  mj,  7712,  m',  etc.,  so 
that  m  =  tan  a,  m^  =  tan  aj,  etc. 

The  inclination  may  be  any  angle 
from  0  to  TT  inclusive.  The  slope  may  be  any  real  number, 
since  the  tangent  of  an  angle  in  the  first  two  quadrants  may  be 
any  number  positive  or  negative.  The  slope  of  a  line  parallel 
to  X'X  is  of  course  zero,  since  the  inclination  is  0  or  tt.  For  a 
line  parallel  to  Y'  Y  the  slope  is  infinite. 

Theorem.     The  slope  m  of  the  line  passing  throHfjh  two  points 
,,.^r.^iQpiy  Vi)^  ^M^2,  2/2)  'is  given  by 


(H) 


Xi  —  X2 


(1) 


Proof.     In  the  figure 

OMi=Xi,  OM2=X2,  MiF^=yj^,  ^¥2^2= 2/2- 

Draw  FojS  parallel  to  OX.  Then  in 
the  right  triangle  P2SP1J  since  angle 
P^PoS  =  a,  we  have 

P,S' 


■  1  an  a  ■■ 


18  ELEMENTARY   ANALYSIS 

But  jSP^  =  M^P^  -  M^S  =  M,P^  -  M2P2  =  2/1  -  ^2 ; 

P2S  =  M^M^  =  OM^  -  OM2  =  x,-  X2. 

Substituting  their  values  in  (1),  gives  (II).    '  q.e.d. 

The  student  should  derive  (II)  when  a  is  obtuse.^ 

We  next  derive  the  conditions  for  parallel  lines  and  for  per- 
pendicular lines,  in  terms  of  their  slopes. 

Theorem.  If  two  lines  are  2')<^'rallel,  their  slopes  are  equal;  if 
perpendicular,  tJie  slope  of  one  is  the  negative  reciprocal  of  the 
slope  of  the  other,  and  conversely. 

Proof  Let  a^  and  ^2  be  the  inclinations  and  m^  and  mg  the 
slopes  of  the  lines. 

If  the  lines  are  parallel,  a^=  a2.     .  •.  m^  =  m^. 
If  the  lines  are  perpendicular,  as  in  the  figure, 

TT   , 
«2  =  o  +  ^1- 


.  m2  =  tan  ccg  =  tan/  7+^1 


.  m2  = Q.E.D. 


=  —  cot  cci  (by  31,  p.  3) 

= ^ —  (by26,  p.  3) 

tan «!  V  J       ^  F     y 

1 

m- 

The  converse  is  proved  by  retracing  the  steps  with  the  as- 
sumption,^in*the  second  part,  that  ^2  is  greater  than  ai. 

The  problem  frequently  arises  :  Given  two  points,  to  find  the- 
coordinates  of  their  middle  point.     This  is  solved  by  means  of 

Theorem.  If  P{x,  y)  is  the  middle  point  of  the  line  ivhose  ex- 
tremities are  Pi{xi,  2/1)  ^^^  -^2(^2?  2/2);  ^^^^^ 

*  To  construct  a  line  passing  through  a  given  point  Pi  whose  slope  is  a 
positive  fraction  - ,  we  mark  a  point  S  b  units  to  the  right  of  Pi  and  a  point 
P2  a  units  above  S,  and  draw  PiPg-  If  the  slope  is  a  negative  fraction, 
—  - ,  then  either  S  must  lie  to  the  left  of  Pj,  or  P2  must  lie  below  S. 


CARTESIAN   COORDINATES 


19 


Y^ 


Pi 


"K      3J       M2JC 


(III)  a^  =  i(xi  +  a^2),  2/  =  K2/i  +  2/2). 

Proof.  Project  the  points  on  OX.  Then,  by  geometry,  since 
PiP  =  *PP2,  then  also 

(1)  3f^M=  MM,. 

Now        OM^  =  x^,    OM=x,    OM2  =  X2. 

Hence    M^M  =x  —  x^,  MM,  =  x,—  x. 

,' ,  X  —  Xi  ^=  x,  —  Xj    or  X  —  -^  (x^  -\-  x,) . 
Similarly,  we  may  show  that 

2/ =  -2- (2/1 +2/2).  Q.E.D. 

PROBLEMS 

1.  Find  the  slope  of  the  line  joining  (1,  3)  and  (2,  7). 

Ans.  4. 

2.  Find  the  slope  of  the  line  joining  (2,  7)  and  (—4,  —4). 

Ans,  -y-. 

3.  Find  the  slope  of  the  line  joining  (V3,  V2)  and 
(-V'2,  V3).  Ans.  2V6-5. 
t^A.   Find    the    slope    of    the    line    joining    (a  +  Z>,    c  +  a), 


(c  -\-  a,  b  -{-  c). 


Ans. 


G~b 


5.  Find  the  slopes  of  the  sides  of  the  triangle  whose 
vertices  are  (1,  1),  (—1,   —1),  (V3,   —  V3). 

Ans.  1,  1±^,  Lz:^. 
1_V3    1+V3 

^  6.   Prove  by  means  of  slopes  that  (—  4,  —  2),  (2,  0),  (8,  6), 
(2,  4)  are  the  vertices  of  a  parallelogram. 

7.  Prove  by  means  of  slopes  that  (3,  0),  (6,  4),  (—1,  3)  are 
the  vertices  of  a  right  triangle. 

8.  Prove  by  means  of  slopes  that  (0,  -  2),  (4,  2),  (0,  6), 
(—4,  2)  are  the  vertices  of  a  rectangle,  and  hence,  by  (I),  of  a 
square. 

9.  Prove  by  means  of  their  slopes  that  the  diagonals  of  the 
square  in  problem  8  are  perpendicular. 


20  ELEMENTARY   ANALYSIS 

10.  Prove  by  means  of  slopes  that  (10,  0),  (5,  5),  (5,  —  5), 
(—  5,  5)  are  the  vertices  of  a  trapezoid. 

11.  Show  that  the  line  joining  (a,  h)  and  (c,  —  d)  is  parallel 
to  the  line  joining  (—a,  —  h)  and  (— c,  d). 

12.  Show  that  the  line  joining  the  origin  to  (a,  h)  is  perpen- 
dicular to  the  line  joining  the  origin  to  {—h,  a). 

13.  What  is  the  inclination  of  a  line  parallel  to  Y'Y?    per- 
pendicular to  Y'Y? 

14.  AYhat  is  the  slope  of  a  line  parallel  to  Y'  Y?    perpen- 
dicular to  Y'Y? 

15.  What  is  the  inclination  of  the  line  joining  (2,  2)  and 

4 

16.  W^hat  is  the  inclination  of  the  line  joining  (—2,  0)  and 

4 

17.  What  is  the  inclination  of  the  line  joining  (3,  0)  and 

3 

/  18.    What  is  the  inclination  of  the  line  joining  (3,  0)  and 

3 

19.  What  is  the  inclination  of  the  line  joining  (0,  —  4)  and 

6 

20.  What  is  the  inclination  of  the  line  joining  (0,  0)  and 
(-V3,  1)?  .  ^        Stt 

6 

21.  Prove  by  means  of  slopes  that  (2,  3),  (1,  —3),  (3,  9)  lie 
on  the  same  straight  line. 

22.  Prove  that  the  points  (a,  &  +  c),  (h,  c  +  ct),  and  (c,  a  +  b) 
lie  on  the  same  straight  line. 

23.  Prove  that  (1,  5)  is  on  the  line  joining  the  points  (0,  2) 
and  (2,  8)  and  is  equidistant  from  them. 


CARTESIAN   COOKDINATES  21 

24.  Prove  that  the  line  joining  (3,  —  2)  and  (5,  1)  is  perpen- 
dicular to  the  line  joining  (10,  0)  and  (13,  —  2). 

25.  Find  the  coordinates  of  the  middle  point  of  the  line 
joining  (4,  —6)  and  (—2,  —4).  Ans.  (1,  —5). 

26.  Find  the  coordinates  of  the  middle  point  of  the  line 
joining  (a  +  b,  c  +  d)  and  (a  —  b,  d  —  c),  Ans.  (a,  d). 

27.  Find  the  middle  points  of  the  sides  of  the  triangle 
whose  vertices  are  (2,  3),  (4,  —5),  and  (—3,  —6);  also  find 
the  lengths  of  the  medians. 

28.  Prove  that  the  middle  point  of  the  hypotenuse  of  a  right 
triangle  is  equidistant  from  the  three  vertices. 

29.  Show  that  the  diagonals  of  the  parallelogram  whose 
vertices  are  (1,  2),  (—5,  —3),  (7,  —6),  (1,  —11)  bisect  each 
other. 

30.  Prove  that  the  diagonals  of  any  parallelogram  mutually 
bisect  each  other. 

31.  Show  that  the  lines  joining  the  middle  points  of  the 
opposite  sides  of  the  quadrilateral  whose  vortiOes  are  (6,  8V 
(-  4,  0),  (-  2,  -  6),  (4,  -  4)  bisect  each  other. 

32.  In  the  quadrilateral  of  problem  31  &how  by  means  of 
slopes  that  the  lines  joining  the  middle  points  of  the  adjacent 
sides  form  a  parallelogram. 

33.  Show  that  in  the  trapezoid  whose  vertices  are  (—8,  0), 
(—  4,  —4),  (—  4,  4),  and  (4,  —  4)  the  length  of  th©  line  joining 
the  middle  points  of  the  non-parallel  sides  is  equal  to  one  half 
the  sum  of  the  lengths  of  the  parallel  sides.  Also  prove  that 
it  is  parallel  to  the  parallel  sides. 

13.  Areas.  In  this  section  the  problem  of  determining  the 
area  of  any  polygon,  the  coordinates  of  whose  vertices  are 
given,  will  be  solved.     We  begin  with 

Theorem.  The  area  of  a  tria.yigle  whose  vertices  are  the  oriyin, 
-f*i(^i7  2/i)?  ^^^  Afe  2/2)  **s  given  by  the  formula 


22 
(IV) 

Y 


ELEMENTARY   ANALYSIS 

Area  of  triangle  OPiPz  =  I  {pciyo  —  oc^Vi)  - 
Po(x,,y,)  Proof,     In  the  figure  let 


M,  JC 


a  =  Z  XOP^, 

/3  =  ZX0F,,  , 

e  =  ZP,OP,,  ^^y-^' 

By  45^4^  C.P  /I   .  i  ^- 1^-- ^ 

(2)  Area  A  OP1P2  =  1  OP^  -  OP,  sin  (9 

=  iOP,.OAsin(^^a)  [by(l)] 

(3)  =  i  OPi .  OP2  (sin  ^  cos  a  -  cos  /?  sin  a). 

(by  34,  p.  3) 
But  in  the  figure 

OP/ 


sin^ 


^^^=-/L-cos^  = 


OP2 

_M,P,_   y. 


OP,      OP, 
0M^_    x^ 


sina= — - — i  =  -!^J— ,  cos  a 

OP^       OP/  OP,      OP, 

Substituting  in  (3)  and  reducing,  we  obtain 
Area  A  OP^P,  =  i  {x,y,  —  x^y,). 

EXAMPLE 


Q.E.D. 


Eind  the  area  of  the  triangle  whose  vertices  are  the  origin, 
(-2,  4),  and  (-5,  -1). 

Solution.     Denote  (  —  2, 4)  by  Pi,  (  -^  5, 
-  1)  by  P2.     Then 

x,  =  —  2,   2/1  =  4,   x2  =  —  5,   y ,■=  —  !. 

Substituting  in  (IV), 

Area  =  1  [_  2  •  - 1  -  (-  5)  •  4]  =  11. 

Then  Area  =  11  \init  squares. 

If,   however,   the    formula    (IV)    is    applied    by   denoting 
(-  2,  4)  by  P2,  and  (-  5,  - 1)  by  Pi,  the  result  will  be  - 11. 


r-2 

A) 

Yk 

\ 

/ 

\ 

i 

/ 

\ 

/ 

\ 

(1,1) 

J 

\ 

5 

-^ 

L 

^ 

^^ 

0 

X 

(-5 

-1) 

CARTESIAN   COORDINATES 


23 


The  two  figures  are  as  follows : 
The  cases  of  positive  and  neg- 
ative area  are  distinguished  by 

Theorem.  Passing  around  the 
perimeter  in  the  order  of  the  ver- 
tices 0,  Fi,  P2, 

if  the  area  is  on  the  left,  as  in  Fig,  1,  then  (IV)  gives  a  posi- 
tive result; 
if  the  area  is  on  the  right,  as  in  Fig,  2,  then  (IV)  gives  a  nega- 
tive result. 

Proof     In  the  formula 

(4)  Area  A  OP^P^  =  i  OP^  -  OP2  sin  0 

the  angle  0  is  measured  fi^om  OP^  to  OP2  within  the  triangle, 

P^  p^  Hence  0  is  positive  when  the  area  lies 

.V^^~~^^^^^II_Pi   ^^^""""-^-^2  to   tlie   left   in   passing   around   the 

^^^X^     \  Y^v-'""''^     perimeter  0,  P^,  P^,  as  in  Fig.  1,  since 

Q  O  ^  is  then  measured  counter-clockwise 

(1)  ^^^  (p.  2).     But  in  Fig.  2,  0  is.  measureJi 

clockwise.     Hence  6  is  negative  and  sin  0  in  (4)  is  alsp  negative. 


S 


Q.E.D. 


Formula  (IV)  is  easily  applied  to  any  polygon  by  regarding 
its  area  as  made  up  of  triangles  with  the  origin  as  a't  common 
vertex.     Consider  any  triangle.  ^ 

Theorem.     The  area  of  a  triangle  whose  vertices  are  P^  {x^,  2/1); 
P2{^2,  2/2),  A (^3,  2/3)  i8  given  by 
(V)   Area  A  P1P2P3  =  i(^i2/2  -  a?22/i  +  i^22/3  -  003y2  +  ocsiji  -  Xiy^). 

This  formula  gives  a  positive  or  negative  result  according  as 
the  area  lies  to  the  left  or  right  in  pass- 
ing  around  the  perimeter  in  the  order 
Pi  P2  P3. 

Proof  Two  cases  must  be  distin- 
guished according  as  the  origin  is 
within  or  without  the  triangle. 


24  .    ELEMENTARY   ANALYSIS 

Fig.  1,  origin  within  the  triangle.     By  inspection, 

(5)  Area  A  P,P,P,  =  A  OP,P,  +  A  OP,P,  +  A  OP,P^, 
since  these  areas  all  have  the  same  sign.  , 

Fig.  2,  origin  imthout  the  triangle.     By  inspection, 

(6)  Area  A  P^P.P^  =  A  OP^Po  +  A  OP2P3  +  A  OPs^i, 

since  OP1P2,  OP3P1  have  the  same  sign,  but  OP^A  the  opposite 
sign,  the  algebraic  sum  giving  the  desired  area. 

By  (IV),    A  OP,P,  =  i  (x,y,  -  x.jj,\ 

A  OP2P3  =  \  {x.y^  -  x^y^),  A  OP3P1  =  ^  {x^y^  -  x,y^. 

Substituting  in  (5)  and  (6),  we  have  (V). 

Also  in  (5)  the  area  is  positive,  in  (6)  negative.  q.e.d. 

An  easy  way  to  apply  (V)  is  given  by  the  following 

Rule  for  finding  the  area  of  a  triangle.  x^  t/i 

First  step.      Write  down  the  vertices  in  two  columns,  X2  2/2 

abscissas  in  one,   ordinates    in   the  other,  repeating  the  x^  y^ 

coordinates  of  the  first  vertex.  x^  y^ 

Second  step.  Multiply  each  abscissa  by  the  ordinate  of  the  next 
row,  and  add  results.     This  gives  x^y^  +  ^^'2^/3  +  ^zVi- 

Third  step.  Multiply  each  ordinate  by  the  abscissa  of  the  next 
row,  and  add  results.     This  gives  yiX2  +  ^2^3  +  2/3^1- 

Fourth  step.  Subtract  the  residt  of  the  third  step  fro7n  that  of 
the  second  step,  and  divide  by  2.  This  gives  the  required  area, 
namely,  formula  (V). 

It  is  easy  to  show  in  the  same  manner  that  the  rule  applies 
to  any  polygon,  if  the  following  caution  be  observed  in  the 
first  step : 

Write  down  the  coordinates  of  the  vertices  in  an  order  agreeing 
with  that  estal)lished  by  passing  continuously  around  the  perime- 
ter,  and  repeat  the  coordinates  of  the  first  vortex. 


CARTESIAN   COORDINATES 


25 


EXAMPLE 

Find  the  area  of  the  quadrilateral  whose  vertices  are  (1,  6), 
(-3,  -4),  (2,  -2),  (-1,3). 

Solution. 


1 
-1 
-3 

2 
1 


6 

3 

-4 

-2 
6 


Plotting,  we  have 
the  figure  from  which  we 
choose  the  order  of  the  ver- 
tices as  indicated  by  the  ar- 
rows.    Following  the  rule : 

First  step.  Write  down  the  vertices  in 
order. 

Second  step.  Multiply  each  abscissa  by 
the  ordinate  of  the  next  row,  e^lj^  add. 
This  gives 

lx3  +  (-lX  -4)4-(-3x  -2)-f2x6  =  25. 

Third  step.     Multiply  each  ordinate  by  the  abscissa  of  the 
next  row,  and  add.     This  gives 

6  X -1+3  X -3  + (-4x2) +  (-2x1)  =  -25. 

Fourth  step.     Subtract  the  result  of  the  third  step  from  the 
result  of  the  second  step,  and  divide  by  2. 

25  4. 25 

.*.  Area  =         -^-  =  25  unit  squares.     Ans. 

The  result  has  the  positive  sign,  since  the  area  is  on  the  left. 

PROBLEMS 

1.  Find  the  area  of  the  triangle  whose  vertices  are  (2,  3), 
(1,  5),  (-1,  -2).  Ans.  -V-. 

2.  Find  the  area  of  the  triangle  whose  vertices  are  (2,  3), 
(4,  -5)  (-3,  -6).  Ans.  29. 

3.  Find  the  area  of  the  triangle  whose  vertices  are  (8,  3), 
(-2,3),  (4,  -5).  Ans.  40. 

4.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  0), 
(- a,  0),  (0,  h).  .  Ans.  ab. 


26  ELEMENTARY   ANALYSIS 

5.  Find  the  area  of  the  triangle  whose  vertices  are  (0,  0), 
(P^h  Vi),  (^2,  2/2)-  Ans,  ^^^2  -  ^2^1 . 

6.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  1), 
(0,6),(c,l).  Ans.  (°^-<^K^-l). 

7.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  &), 
(h,  a),  (c,  —  c).  Ans.  \  (or  —  W). 

8.  Find  the  area  of  the  triangle  whose  vertices  are  (3,  0), 
(0,  3  V3),  (6,  3  V3).  Ans,  9  V3. 

9.  Prove  that  the  area  of  the  triangle  whose  vertices  are 
the  points  (2,  3),  (5,  4),  (-^*^lj  is  zero,  and  hence  that  these 
points  all  lie  on  the  same  straight  line. 

10.  Prove  that  the  area  of  the  triangle  whose  vertices  are 
the  points  (a,  h  +  c),  (b,  c  +  a),  (c,  a-{-b)  is  zero,  and  hence 
that  these  points  all  lie  on  the  same  straight  line. 

11.  Prove  that  the  area  of  the  triangle  whose  vertices  are 
the  points  (a,  G-]-a),  (— c,  0),  (—  a,  c  —  a)  is  zero,  and  hence 
that  these  points  all  lie  on  the  same  straight  line. 

12.  Find  the  area  of  the  quadrilateral  whose  vertices  are 
(-2,  3),  (-3,  -4),  (5,  -1),  (2,  2).  Ans.  31. 

13.  Find  the  area  of  the  pentagon  whose  vertices  are  (1,  2), 

■  (3,  - 1),  (6,  -  2),  (2,  5),  (4,  4).  ,  Ans.  18. 

14.  Find  the  area  of  the  parallelogram  whose  vertices  are 
(10,  5),  (-2,  5),  (-5,  -3),  (7,  -3).  Ans.  96. 

15.  Find  the  area  of  the  quadrilateral  whose  vertices  are 
(0,  0),  (5,  0),  (9,  11),  (0,  3).  Ans.  41. 

'  16.   Find  the  area  of  the  quadrilateral  whose  vertices  are 
(7,  0),  (11,  9),  (0,  5),  (0,  0).  Ans.  59. 

17.  Show  that  the  area  of  the  triangle  whose  vertices  are 
(4,  6),  (2,  —4),  (—4,  2)  is  four  times  the  area  of  the  triangle 
formed  by  joining  the  middle  points  of  the  sides. 


CARTESIAN  COORDINATES  *  27 

18.  Show  that  the  lines  drawn  from  the  vertices  (3,  —  8), 
(__  4^  6),  (7,  0)  to  the  medial  point  of  the  triangle  divide  it 
into  three  triangles  of  equal  area. 

19.  Given  the  quadrilateral  whose  vertices  are  (0,  0),  (6,  8), 
(10,  —  2),  (4,  —  4)  ;  show  that  the  area  of  the  quadrilateral 
formed  by  joining  the  middle  points  of  its  adjacent  sides  is 
equal  to  one  half  the  area  of  the  given  quadrilateral. 


CHAPTER   111 

CURVE  AND  EQUATION 

14.  Locus  of  a  point  satisfying  a  given  condition.    The 

curve  ^  (or  group  of  curves)  passing  through  all  points  which 
satisfy  a  given  condition,  and  through  no  other  points,  is  called 
the  locus  of  the  point  satisfying  that  condition. 

For  example,  in  Plane  Geometry,  the  following  results  are 
proved : 

The  perpendicular  bisector  of  the  line  joining  two  fixed 
points  is  the  locus  of  all  points  equidistant  from  these  ]3oints. 

The  bisectors  of  the  adjac^t  angles  formed  by  two  lines  are* 
the  locus  of  all  points  equidistant  from  these  lines. 

To  solve  any  locus  problem  involves  two  things : 

1.  To  draw  the  locus  by  constructing  a  sufficient  number  of 
points  satisfying  the  given  conditio^  and  therefore  lying  on 
the  locus. 

2.  To  discuss  the  nature  of  the  locus,  that  is,  to  determine 
properties  of  the  curve. 

Analytic  Geometry  is  peculiarly  adapted  to  the  solution  of 
i^pth  parts  of  a  locus  problem. , 

15.  Equation  of  the  locus  of  a  point  satisfying  a  given 
condition.  Let  us  take  up  the  locus  problem,  making  use  of 
coordinates.  We  imagine  the  point  P(x,  y)  moving  in  such  a 
manner  that  the  given  condition  is  fulfilled.  Then  the  given 
condition  will  lead  to  an  equation  involving  the  variables  x 
and  ?/.     The  following  example  illustrates  this. 

*  The  word  "curve"  will  hereafter  signify  anij  continuous  line,  straight 
or  curved. 

28 


CURVE   AND   EQUATION 


29 


EXAMPLE 

The  point  P(x,  y)  moves  so  that  it  is  always  equidistant  from 
j^(_  2,  0)  and  B{—3,  8).     Find  the  equation  of  the  locus. 

Solution.     Let  P{x,  y)  be  any  point  on  the  locus.     Then  by 
the  given  condition 

(1)     FA  =  FB. 

But,  by  formula  (I),  p.  13, 


FA  =  -\/{x  +  2y  -h  (2/  -  0)2, 


FB  =  V(x-{-3y-\-{y- 
Substituting  in  (1), 

(2)     ^(x  +  2y-{-(y-0y 


■  sy 


=^(x-^3y+(y-sy. 

Squaring  and  reducing, 

(3)  2x-16y  +  69  =  0, 

In  the  equation  (3),  x  and  y  are  variables  representing  the 
coordinates  of  any  point  on  the  locus;  that  is,  of  any  point 
on  the  perpendicular  bisector  of  the  line  AB.  This  equation 
is  called  the  equation  of  the  locus ;  that  is,  it  is  the  equation  of 
the  perpendicular  bisector  CF.  It  has  two  important  and 
characteristic  properties : 

1.  The  coordinates  of  any  point  on  the  locus  may  be  sub- 
stituted for  X  and  y  in  the  equation  (3),  and  the  result  will  be 
true. 

For  let  Pi  (xi,  y^  be  any  point  on  the  locus.  Then  F^A  =  F^B, 
by  definition.     Hence,  by  formula  (I),  p.  13, 

(4)  VK  +  2y  +  y,'  =  V(^i  +  3)2  +  (2/1^=^ 
or,  squaring  and  reducing, 

(5)  2a;i-16?/i  +  69  =  0. 

But  this  eqaiation  is  obtained  by  substituting  x^  and  .Vi  foi"  ^ 
and  y,  respectively,  in  (3).     Therefore  x^  and  y^  satisfy  (3). 


30  ELEMENTARY  ANALYSIS 

2.  Conversely,  every  point  whose  coordinates  satisfy  (3)  will 
lie  upon  the  locus. 

For  if  Pi  (xi,  ?/i)  is  a  point  whose  coordinates  satisfy  (3),  then 
(5)  is  true,  and  hence  also  (4)  holds.  q.e.d. 

In  particular,  the  coordinates  of  the  middle  point  0  of  A 
and  Bj  namely,  a;  =  —  2^,  ?/  =  4  (III,  p.  19),  satisfy  (3),  since 

2(_2-i-)-16x  4  +  69  =  0. 

This  discussion  leads  to  the  definition : 

The  equation  of  the  locus  of  a  point  satisfying  a  given  condi- 
tion is  an  equation  in  the  variables  x  and  y  representing  coor- 
dinates such  that  (1)  the  coordinates  of  every  point  on  the 
locus  will  satisfy  the  equation ;  and  (2)  conversely,  every  point 
whose  coordinates  satisfy  the  equation  will  lie  upon  the  locus. 

This  definition  shows  that  the  equation  of  the  locus  must  be 
tested  in  two  ways  after  derivation,  as  illustrated  in  the  exam- 
ple of  this  section.  The  student  should  supply  this  test  in  the 
examples,  p.  31,  and  problems,  p.  32. 

Erom  the  above  definition  follows  at  once  the 

Corollary.  A  point  lies  upon  a  curve  when  and  only  when  its 
coordinates  satisfy  the  equation  of  the  curve. 

16.  First  fundamental  problem.  To  find  the  equation  of  a 
curve  which  is  defined  as  the  locus  of  a  point  satisfying  a  given 
condition. 

The  following  rule  will  suffice  for  the  solution  of  this  prob- 
lem in  many  cases : 

Rule.  First  step.  Assume  that  P{x,  y)  is  any  point  satisfying 
the  given  condition  and  is  therefore  on  the  curve. 

Second  step.      Write  down  the  given  condition. 

Third  step.  Express  the  given  condition  in  coordinates,  and 
simplify  the  result.  The  final  equation,  containing  x,  y,  and  the 
given  constants  of  the  problem,  will  be  the  required  equation. 


CURVE   AND   EQUATION 


31 


EXAMPLES 

1.    Find  the  equation  of  the  straight  line  passing  through 

Sir 
Pj  (4^  —  1)  and  having  an  inclination  of 

Solution.     First  step.     Assume  P(xj  y)  any  point  on  the  line. 
Second  step.     The  given  condition,  since 


the  inclination  a  is  — ,  may  be  written 

(1)  Slope  of  PiP  =  tan  a  =  - 1. 
Tliird  step.     From  (II),  p.  IT, 

(2)  Slopeof  PiP=  tan  «  =  -^^^-^:i^2  =  -^^^  - 
[By  substituting  {x,  y)  for  (a;,,  y^,  and  (4,  —  1)  for  (a;^,  y^."] 


y{ 

^ 

^: 

\ 

r 

'\ 

V  i 

\ 

it. 

J 

\ 

0 

\ 

a,- 

■1) 

— 

— 

i-.S 

.'.  from  (1), 


x  —  4z 


1, 


or 
(3) 


x-{-y  —  3  =  0.     Ans, 


2.  Find  the  equation  of  a  straight  line  parallel  to  the  axis  of 
y  and  at  a  distance  of  6  units  to  the 
rierht. 


f^ 


M 


Solution.      First   step.     Assume   that 
^-^  P(x,  y)  is  any  point  on  the    line,  and 
draw  NP  perpendicular  to  0 1^ 

Second   step.      The    given    condition 
►^    may  be  written 


(4)    NP  =  6. 

Third  step.     Since  NP=  OM==x,  (4)  becomes  ^^ 

(5)  x  =  6.     Ans. 

3.    Find  the  equation  of  the  locus  of  a  point  whose  distance 
from  (—1,  2)  is  always  equal  to  4. 


32 


ELEMENTARY   ANALYSIS 


Solution.     First  step.     Assume  that  P{x,  y)  is  any  point  on 

the  locus. 

Second  step.     Denoting  ( —  1^  2) 
)     by  C,  the  given  condition  is 

(6)  Pe=:4. 

TJiird  step.     By  formula  (I),  p. 
^.     13, 


PG  =  -V(x-\-iy-i-(y-2f. 
Substituting  in  (5), 


-V(x+iy  +  (j/-2y=4.. 

Squaring  and  reducing, 

(7)  x^  +  y^  +  2x-^y-ll==0. 

This  is  the  required  equation,  namely,  the  equation  of  the 
circle  whose  center  is  (—1,  2)  and  radius  equal  4. 


'\t^. 


PROBLEMS 

1.  Find  the  equation  of  a  line  parallel  to  OF  and. 

(a)  at  a  distance  of  4  units  to  the  right. 

(b)  at  a  distance  of  7  units  to  the  left. 

(c)  at  a  distance  of  2  units  to  the  right  of  (3,  2). 

(d)  at  a  distance  of  5  units  to  the  left  of  (2,  —  2).  >l  •  *  ^ 

2.  Find  the  equation  of  a  line  parallel  to  OX  and 

(a)  at  a  distance  of  3  units  above  OX. 

(b)  at  a  distance  of  6  units  below  OX. 

(c)  at  a  distance  of  7  units  above  (—2,-3). 

(d)  at  a  distance  of  5  units  below  ^4,  —  2). 

3.  What  is  the  equation  of  XX'  ?  of  YY'?  %^' ^  ' 

4.  Find  the  equation  of  a  line  parallel  to  the  line  x  =  4z  and 
3  units  to  the  right  of  it.     Eight  units  to  the  left  of  it. 

5.  Find  the  equation  of  a  line  parallel  to  the  line  y  =  —  2 
and  4  units  below  it.     Five  units  above  it. 


CUKVE   AND   EQUATION  33 

6.  What  is  the  equation  of  the  locus  of  a  point  which  moves 
always  at  a  distance  of  2  units  from  the  axis  of  x  ?  from  the 
axis  of  ?/  ?    from  the  line  ic  =  —  5  ?    from  the  line  2/  =  4  ? 

7.  What  is  the  equation  of  the  locus  of  a  point  which  moves 
so  as  to  be  equidistant  from  the  lines  a;  =  5  and  a;  =  9  ?  equi- 
distant from  ^  =  3  and  ?/  =  —  7  ? 

8.  What  are  the  equations  of  the  sides  of  the  rectangle 
whose  vertices  are  (5,  2),  (5,  5),  (-2,  2),  (-2,  5)? 

In  problems  9  and  10,  P^  is  a  given  point  on  the  required  line, 
m  is  the  slope  of  the  line,  and  a  its  inclination. 

9.  What  is  the  equation  of  a  line  if 

(a)   Fi  is  (0,  3)  and  m  =  -  3  ?  ^ns.  Z  x-\-y -?>=.^, 

^(6)   Pi  is  (-  4,  -  2)  and  m  =  |  ?     Ans.  x-3y-2  =  0. 

(c)    Pi  is  (-2,  3)  and  m=^?      ^ns.   -s/2x-2yf6 

J  _     +2V2  =  0. 

((^)  Pi  is  (0,  5)  and  711  =  ^?  Ans.  V3 a;-2 ?/  +  10  =  0. 

/ 

"^  (e)  Piis  (0,  0)  andm  =  -|?  ^ns.  2a^4-3?/  =  0. 

(/)  Pi  is  (a,  5)  and  m  =  0  ?  ^tis.  ?/  =  &. 

(9')  Pi  is  (—  a,  Z>)  and  771  =  00  ?  ^dliis.  x  =  —  a. 

10.    What  is  the  equation  of  a  line  if 

^  (a)  Pi  is  (2,  3)  and  a  =  45°?  Ans.  x-y-{-l  =  0. 

^  (b)  Pi  is  (- 1,  2)  and  r^e  =  45° ?  ylns.  a;  -  2/  +  3  =  0. 

(c)  Pi  is  (—  a,  —  6)  and  a  =  45°  ?  Ans.  x  —  y  =  b  —  a. 

(d)  Pi  is  (5,  2)  and  r^  =  60°  ?  ^ris.  ■\/3x-y  +  2 

-5V3  =  0. 

(e)  Pi  is  (0,  -  7)  and  a  -  60°  ?        ^ns.    VSx-y~7  =  0. 
(/)  Pi  is  (-  4,  5)  and  a  =  0°?  Ans.  y  =  5. 

(g)  Pi  is  (2,  -  3)  and  a  =  90°  ?  A71S.  x  =  2. 

Qi)  Piis(3,-3V3)anda:=rl20°?  Ans.  ^3x^y  =  0, 

(i)  Pi  is  (0,  3)  and  a  -  150°  ?  Ans.   VS  x-i-S  y -9  =0. 

(j)  Pi  is  (a,  b)  and  a  =  135°  ?  ^ns.  a;  +  2/  =  a  +  6. 


34  ELEMENTARY  ANALYSIS 

11.  Find  the  equation  of  the  circle  with 

(a)  center  at  (3,  2)  and  radius  =  4. 

Arts,  x^ -\- y^  —  6 X  —  4:y  —  3  =  0, 

(b)  center  at  (12,  —5)  and  r=13. 

Aris.  x^  +  y'--24.x  +  10y  =  0, 

(c)  center  at  (0,  0)  and  radius  =  r,  Ans.  x^-\-y^  =  rl 

(d)  center  at  (0,  0)  and  r  =  5.  Ans.  x^-\-y^  =  25. 

(e)  center  at  (3  a,  4  a)  and  r  =  5  a. 

Ans.  x^  -\- y^  —  2  a{3  X  +  4:  y)=  0. 
(/)  center  at  (6  +  c,  b  —  c)  and  r  =  c. 

.4ns.  a;-  +  2/'  -  2(^  +  c)a:  -2(b-c)y  +  2b^  +  c^  =  0. 

12.  Find  the  equation  of  a  circle  whose  center  is  (5,  —  4) 
and  whose  circumference  passes  through  the  point  (—2,  3). 

13.  Find  the  equation  of  a  circle  having  the  line  joining 
(3,  —  5)  and  (—  2,  2)  as  a  diameter. 

14.  Find  the  equation  of  a  circle  touching  each  axis  at  a  dis- 
tance 6  units  from  the  origin. 

15.  Find  the  equation  of  a  circle  whose  center  is  the  middle 
point  of  the  line  joining  (—6,  8)  to  the  origin  and  whose  cir- 
cumference passes  through  the  point  (2,  3) . 

16.  A  point  moves  so  that  its  distances  from  the  two  fixed 
points  (2,  —3)  and  (—1,  4)  are  equal.  Find  the  equation  of 
the  locus.     Ans.     3x  —  7y-\-2  =  0. 

17.  Find  the  equation  of  the  perpendicular  bisector  of  the 
line  joining 

.    (a)  (2,1),  (-3/ -3).  Ans.  10  a; +8^  +  13  =  0. 

(b)  (3,  1),  (2,  4).  Ans.  x-3y-{-5  =  0. 

(c)  (-1^  _1)^  (3,  7).  Ans.  x  +  2y-7=:0. 

(d)  (0,  4),  (3,  0).  Ans.  6x-Sy  +  7  =  0, 

(e)  (xi,  ?/i),  (X2,  2/2). 

Ans.    2  (x^  —  0^2)  ^  +  2  (^1  -  2/2)2/  +  ^2^  —  ^i  +  ^2'  —  Vi^  =  0. 

18.  Show  that  in  problem  17  the  coordinates  of  the  middle 
point  of  the  line  joining  the  given  points  satisfy  the  equation 
of  the  perpendicular  bisector. 


I 


CURVE  AND  EQUATION  35 


9.  Find  the  equations  of  the  perpendicular  bisectors  of  the 
sides  of  the  triangle  (4,  8),  (10,  0),  (6, 2).  Show  that  they  meet 
in  the  point  (11,  7). 

17.  Locus  of  an  equation.  The  preceding  sections  have 
illustrated  the  fact  that  a  locus  problem  in  Analytic  Geometry 
leads  at  once  to  an  equation  in  the  variables  x  and  y.  This 
equation  having  been  found  or  being  given,  the  complete  solu- 
tion of  the  locus  problem  requires  two  things,  as  already  noted 
in  the  first  section  (p.  28)  of  this  chapter,  namely, 

1.  To  draw  the  locus  by  plotting  a  sufficient  number  of  points 
whose  coordinates  satisfy  the  given  equation,  and  through 
which  the  locus  therefore  passes. 

2.  To  discuss  the  nature  of  the  locus,  that  is,  to  determine 
properties  of  the  curve. 

These  two  problems  are  respectively  called : 

1.  Plotting  the  locus  of  an  equation  (second  fundamental 
problem). 

2.  Discussing  an  equation  (third  fundamental  problem). 

For  the  present,  then,^  we  concentrate  our  attention  upon 
some  given  equation  in  the  variables  x  and  y  (one  or  both)  and 
start  out  with  the  definition : 

The  locus  of  an  equation  in  two  variables  representing  co- 
ordinates is  the  curve  or  group  of  curves  passing  through  all 
points  whose  coordinates  satisfy  that  equation,^*  and  through 
such  points  only. 

*  An  equation  in  the  variables  x  and  y  is  not  necessarily  satisfied  by  the 
coordinates  of  any  points.  For  coordinates  are  real  numbers,  and  the  form 
of  the  equation  may  be  such  that  it  is  satisfied  by  no  real  values  of  x  and  y. 
For  example,  the  equation 

iB2  +  ?/2  +  l  =  0 

is  of  this  sort,  since,  when  x  and  ?/  are  real  numbers,  cc^  and  ?/2  are  necessarily 
positive  (or  zero),  and  consequently  x^  +  y'^-\-l  is  always  a  positive  number 
greater  than  or  equal  to  1,  and  therefore  not  equal  to  zero.  Such  an  equation 
therefore  has  no  locus.  The  expression  "  the  locus  of  the  equation  is  imagi- 
nary "  is  also  used. 

An  equation  may  be  satisfied  by  the  coordinates  of  2i  finite  number  of  points 


36  ELEMENTARY  ANALYSIS 

From  this  definition  the  truth  of  the  following  theorem  is  at 
once  apparent : 

Theorem  I.  If  the  form  of  the  given  equation  be  changed  in  any 
way  {for  example^  by  transposition,  by  multiplication  by  a  constant, 
etc.),  the  locus  is  entirely  unaffected. 

We  now  take  up  in  order  the  solution  of  the  second  and  third 
fundamental  problems. 

18.   Second  fundamental  problem. 

Rule  to  plot  the  locus  of  a  given  equation. 

First  step.  Solve  the  given  equation  for  one  of  the  variables 
in  terms  of  the  other.* 

Second  step.  By  this  formula  compute  the  values  of  the  vari- 
able for  tvhich  the  equation  has  been  solved  by  assuming  real 
values  for  the  other  variable. 

Third  step.  Plot  the  points  corresponding  to  the  values  so 
determined,  t 

Fourth  step.  If  the  points  are  7iumerous  enough  to  suggest 
the  general  shape  of  the  locus,  draw  a  smooth  curve  through  the 
points. 

Since  there  is  no  limit  to  the  number  of  points  which  may  be 
computed  in  this  way,  it  is  evident  that  the  locus  may  be  drawn 
as  accurately  as  may  be  desired  by  simply  plotting  a  sufficiently 
large  number  of  points. 

Several  examples  will  now  be  worked  out.  The  arrangement 
of  the  work  should  be  carefully  noted. 

only.  For  example,  q;2  +  ?/2  =  o  is  satisfied  by  a;  =  0,  ?/  =  0,  but  by  no  other  real 
values.  In  this  case  the  group  of  points,  one  or  more,  whose  coordinates 
satisfy  the  equation,  is  called  the  locus  of  the  equation. 

*  The  form  of  the  given  equation  will  often  be  such  that  solving  for  one 
variable  is  simpler  than  solving  for  the  other.  Alvmys  choose  the  simpler 
solutio7i. 

t  Remember  that  real  values  only  may  be  used  as  coordinates. 


CURVE  AND  EQUATION 


87 


\YJk 

^. 

Y 

\y 

/ 

Y 

^ 

y^ 

/ 

Y 

(0.2) 

A 

r 

r 

(-3 

0) 

0 

X 

if 

EXAMPLES 

1.  Draw  the  locus  of  the  equation 

Solution.     First  step.     Solving  for  ?/, 

?/  =  I  a;  +  2. 

Second  step.     Assume  values  for  x  and  compute  y,  arranging 
results  in  the  form : 

Thus,  if 

a;  =  l,2/  =  | -1  +  2  =  21, 
aj  =  2,2/  =  |-2  +  2  =  3i 
etc. 
Tliird    step.      Plot    the    points 
found. 

Foiirth   step.     Draw    a    smooth 
curve  through  these  points. 

2.  Plot  the  locus  of  the  equation 

y  =  x^  —  2  X  —  3. 

Solution.     First  step:     The  equation  as  given  is  solved  for  y. 
Second  step.     Computing  y  by  assuming  values  of  Xj  we  find 
the  table  of  values  below : 


X 

y 

X 

y 

0 

2 

0 

2 

1 

2! 

-  1 

^ 

2 

H 

-2 

i 

3 

4 

-  3 

0 

4 

4* 

-4 

-! 

etc. 

etc. 

etc. 

etc. 

X 

2/ 

X 

2/ 

0 

o 
—  o 

0 

-3 

1 

-4 

-  1 

0 

2 

-3 

-  2 

5 

3 

0 

-3 

12 

4 

5 

-4 

21 

5 

12 

etc. 

etc. 

6 

21 

etc. 

etc. 

38 


ELEMENTARY  ANALYSIS 


Third  step.     Plot  the  points. 

Fourth  step.     Draw  a  smooth  curve  through  these  points. 
This  gives  the  curve  of  the  figure. 

■^    3.   Plot  the  locus  of  the  equation 

x^  +  y^  +  6x-16  =  0. 

Solution.     First  step.     Solving  for  y, 


y=±  Vl6  —  Qx  —  x^, 
Second  step.     Compute  y  by  assuming  values  of  x. 


X 

y 

X 

2/ 

0 

±4 

0 

±4  . 

1 

±3 

- 1 

±  4.0 

2 

0 

-2 

±  4.9 

3 

imag. 

-3 

±5 

4 

u 

-4 

±  4.9 

5 

u 

-  5 

±4.6 

6 

u 

-6 

±4 

7 

ii. 

—  7 

±3 

-8 

0 

-9 

imag. 

yJ^l^OY  example,  ii  x=%  y  =  ±  Vl6  —  6  —  1  =  ±  3 ; 

ifx  =  3,y=±VW- 
an  imaginary  number ; 


18-9=  ±V-11, 


if  aj  =  - 1,  ?/  =  ±  Vl6  +  6  - 1  =  ±  4.6,  etc. 
TJiird  step.     Plot  the  corresponding  points. 
Fourth  step.     Draw  a  smooth  curve  through  these  points. 

The  student  will  doubtless  remark  that  the  locus  of  example 
1,  p.  37,  appears  to  be  a  straight  line,  and  also  that  the  locus 
of  example  3  (above)  appears  to  be  a  circle.  This  is,  in  fact, 
the  case.     But  the  proof  must  be  reserved  for  later  sections. 


CURVE   AND   EQUATION  39 

PROBLEMS. 

1.    Plot  the  locus  of  each,  of  the  following  equations. 

(a)  x-i-  2  y  —  0.  (m)  y  =  x^  —  x. 


-  x^ 


(b)  x-^2  y  =  3.  ,(n)  y  =  x' 

(c)  3x-y+  5  =  0.  (o)  x^  +  ?r  =  4. 

(d)  y  =  4.x\  ■   {p)  x''-\-\f  =  ^. 
\e)  a;2  +  4?/  =  0.  (q)  x'-{-y'  =  25. 

(/)  y  =  x'-3,  (r)  x'  +  f-^9x=0. 

(g)  x'  +  4.y-5  =  0.  (s)  x'-{-y'-{-4.y  =  0. 

(h)  y^x'  +  x-i-l.  (t)  x''  +  y^-6x-16  =  0. 

(i)  x  =  y^^2y-3.  (u)  x"  ^y'' -  Q>  y -1(5  =  0, 

(/)  4,x  =  y^,                      '  .     (y)  4.y  =  x'^  —  S. 

\t)  y=x^-i,    CjygjQ^ 

2.  Show  that  the  loilowing  equations  have  no  locus  (footnote 
p.  35). 

(a)  x^  +  y'-{-l  =  0.  (e)  (x  +  lf +  f- -\- 4.  =  0, 

(b)  2x'-t3f  =  r^S.  (/)  x'  +  f  +  2x  +  2y  +  S=0. 

(c)  a;2  +  4  =  0.  (g)  4.x' +  7f-i-S  x-^5  =  0. 

(d)  x^-\-y'  +  S  =  0,  (Ji)  y^  +  2x'  +  4.  =  0. 

(i)  9x'  +  Ay'-]-lSx-\-Sy  +  15  =  0. 

Hint.  Write  each  equation  in  the  form  of  a  sum  of  squares,  and  reason  as 
in  the  footnote  on  p.  35. 

The  following  problems  illustrate  the 

Theorem.  If  an  equation  can  be  put  in  the  form  of  a  product 
of  variable  factoids  equal  to  zero,  the  locus  is  found  by  setting  each 
factor  equal  to  zero  and  plotting  each  equation  separately. 

3.  Draw  the  locus  of        4  a;^  —  9  ^^  =  0. 
Solution.    Factoring, 

(1)  C2x-3y){2x-\-3y)  =  0, 

Then,  by  the  theorem,  the  locus  consists  of  the  straight  lines 

(2)  2x~5y  =  0, 

(3)  2x  +  2,y  =  0. 


40  ELEMENTAEY   ANALYSIS 

Proof.  1.  The  coordinates  of  any  point  {xi,  y^  which  satisfy 
(1)  ivill  satisfy  either  (2)  or  (3) . 

For  if  (xi^  2/i)  satisfies  (1), 

(4)  {2x,~3y,){2x,  +  Sy,)=0. 

This  product  can  vanisli  only  when  one  of  the  factors  is 
zero.     Hence  either 

2x,-3y,  =  0, 

and  therefore  (o^j,  2/1)  satisfies  (2) ; 

or  20^1+32/1  =  0, 

and  therefore  (x^,  2/1)  satisfies  (3). 

2.  A  point  (xi,  2/1)  on  either  of  the  lines  defined  by  (2)  and  (3) 
will  also  lie  on  the  locus  of  (1). 

For  if  (x^,  2/1)  is  on  the  line  2x  —  3y  =  0y 
then  (Corollary,  p.  30) 

(5)  2o^.,-3y,  =  0. 

Hence  the  product  (2  iCi  —  3  yi)(2  Xi-\-3  y^)  also  vanishes, 
since  by  (5)  the  first  factor  is  zero,  and  therefore  (xi,  y^  satis- 
fies (1). 

Therefore  every  point  on  the  locus  of  (1)  is  also  on  the  locus 
of  (2)  and  (3),  and  conversely.  This  proves  the  theorem  for 
this  example.  q.e.d. 

4.  Show  that  the  locus  of  each  of  the  following  equations  is 
a  pair  of  straight  lines,  and  plot  the  lines. 

{a)  x'-f^O.  (f)  f-5xy  +  6y  =  0. 

(h)  9i»2-2/2  =  0.  (^)  xy~2x^--3x  =  {), 

(c)  x'  =  <^y\  (h)  xy-2x  =  0. 

(c^)  x^-4:X-5  =  0.  (i)  xy  =  0. 

(e)  y^-6y^7. 

(j  )  3  x^  -\-  o'jj  —  2  y^  -i-  6  X  —  4:  y  =  0. 
(k)  x'^  —  y^  +  x-}-y  =  0,  (m)  x^ -2xy+y^-{-6x-6  y=0. 

(I)  x^-xy~h^x-5y  =  0.  (n)  x^ -iy^ -^  5x -^10 y  =  0, 

(0)  i«2  +  4  i»^/  +  4  /  +  5  a;  +  10  ?/  +  6  =  0. 


CURVE   AND   EQUATION  41 

(q)  x^*—  4,  xy  —  5  y^  -]-  2  X  —  10  y  =  0, 

(r)  Sa^— 2xy  —  y^-{-5x  —  5y  =  0. 

(s)  x^-3xy-4.y'-  =  0. 

(t)    x'  +  2xy-j-y'  +  x-\-y  =  0.        . 

(li)  of —  3  xy  =  0. 

(v)  y^  +  4:  xy  =  0. 

5.  Show  that  the  locus  of  Ax^  +  Bx  +  C  =  0  is  a  pair  of 
parallel  lines,  a  single  line,  or  that  there  is  no  locus  accord- 
ing as  A  =  B^—  4  AG  is  positive,  zero,  or  negative. 

6.  Show  that  the  locus  of  Aoiy^  +  Bxy  +  Gy^  =  0  is  a  pair  of 
intersecting  lines,  a  single  line,  or  a  point  according  as 
A  =  B^  —  4,  AG  is  positive,  zero,  or  negative. 

19.  Third  fundamental  problem.  Discussion  of  an  equa- 
tion. The  method  explained  of  solving  the  second  funda- 
mental problem  gives  no  knowledge  of  the  required  curve 
except  that  it  passes  through  all  the  points  whose  coordinates 
are  determined  as  satisfying  the  given  equation.  Joining 
these  points  gives  a  curve  more  or  less  like  the  exact  locus. 
Serious  errors  may  be  made  in  this  way,  however,  since  tJie 
nature  of  the  curve  between  any  two  successive  points  plotted  is 
not  determined.  This  objection  is  somewhat  obviated  by  de- 
termining before  plotting  certain  properties  of  the  locus  by  a 
discussion  of  the  given  equation  now  to  be  explained. 

The  nature  and  properties  of  a  locus  depend  upon  the  form 
of  its  equation,  and  hence  the  steps  of  any  discussion  must 
depend  upon  the  particular  problem.  In  every  case,  however, 
the  following  questions  should  be  answered. 

1.  Is  the  curve  a  closed  curve,  or  does  it  extend-out  infinitely 
far? 

2.  Is  the  curve  symmetrical  with  respect  to  either  axis  or  the 
origin  ? 

The  method  of  deciding  these  questions  is  illustrated  in  the 
following  examples. 


42 


eleme:ntary  analysis 


EXAMPLES 

1.    Plot  and  discuss  the  locus  of 

(1)  x'  +  4:y'  =  16. 

Solution.   First  step.    Solving  for  x,      -y  -     ^^  /  l^ 

(2)  i»=±2V4^. 

Second  step.     Assume  values  of  y  and  compute  x. 

Third  step.     Plot  the  points  of  the  table. 

Fourth  step.     Draw  a  smooth"  curve  through  these  points. 


X 

y 

X 

y 

±4 

0 

±4 

0 

±3.4 

1 

±3.4 

-1 

±2.1 

li 

±2.7 

-1| 

0 

2 

0 

-2 

imag. 

3 

imag. 

-3 

Discussion.  1.  Equation  (1)  shows  that  neither  x  nor  y  can 
be  indefinitely  great,  since  x^  and  4  2/^  are  positive  for  all  real 
values,  and  their  sum  must  equal  16.  Therefore  neither  x^  nor 
4  y'^  can  exceed  16.     Hence  the  curve  is  a  closed  curve. 

A  second  way  of  proving  this  is  the  following : 

From  (2),  the  ordinate  y  cannot  exceed  2  nor  be  less  than  —  2, 
since  the  expression  4  —  2/^  beneath  the  radical  must  not  be 
negative.  (2)  also  shows  that  x  has  values  only  from  —  4  to 
4  inclusive. 

2.  To  determine  the  symmetry  with  respect  to  the  axes  we 
proceed  as  follows : 

The  equation  (1)  contains  no  odd  powers  of  x  oy  y\  hence  it 
may  be  written  in  any  one  of  the  forms 

(3)  {xf  +  ■4(-  2/)2  =  16,  replacing  {x,  y)  by  {x,  -y)  ; 

(4)  (-  xy  +  4(?/)2  =  16,  replacing  (x,  ?/)  by  (-  x,  y) ; 

(5)  (-  xf  +  4(-?/)2  =  16,  replacing  {x,  y)  by  (-  x,  -y). 


CURVE   AND   EQUATION 


43 


The  transformation  of  (1)  into  (3)  corresponds  in  the  figure 
to  replacing  each  point  P(x,  y)  on  the  curve  by  the  point 
Q(Xy  —y)'  But  the  points  P  and  Q  are  symmetrical  with 
respect  to  XX\  and  (1)  and  (3)  have  the  same  locus  (Theorem 
I,  p.  36).  Hence  the  locus  of  (1)  is  unchanged  if  each  point 
is  changed  to  a  second  point  symmetrical  to  the  first  with 
respect  to  XX\  Therefore  the  locus  is  symmetrical  with  respect 
to  the  axis  of^c^  Similarly,  from  (4),  the  locus  is  symmetrical 
with  respect  to  me  axis  of  y,  and  from  (5),  the  locus  is  symmetri- 
cal with  respect  to  the  origin. 

The  locus  is  called  an  ellipse. 

2.    Plot  the  locus  of 

(6)  2/'-4^  +  15  =  0. 
Discuss  the  equation. 

Solution.  First  step.  Solve  the  equation  for  x,  since  a  square 
root  would  have  to  be  extracted  if  we  solved  for  y.     This  gives 

(7)  x  =  \{f  +  15). 


X 

y 

3| 

0 

4 

±1 

4| 

±2 

6 

±3 

7| 

±4 

10 

±5 

12f 

dz6 

etc. 

etc. 

Second  step.     Assume  values  for  ?/  and  compute  x. 

Since  ?/^  only  appears  in  the  equation,  positive  and  negative 
values  of  y  give  the  same  value  of  x.  The  calculation  gives 
the  table. 


44 


ELEMENTARY   ANALYSIS 


For  example,  ii      y  =  ±  S, 

then  x=:  1(9 +  15)  =  6,  etc. 

TJiii^d  step.     Plot  the  points  of  the  table. 

Fourth  step.     Draw  a  smooth  curve  through  these  points.     , 

Discussion.  1.  From  (7)  it  is  evident  that  x  increases  as  y 
increases.  Hence  the  curve  extends  out  indefinitely  far  from  both 
axes. 

2.  Since  (6)  contains  no  odd  powers  of  y,  the  equation  may 
be  written  in  the  form 

by  replacing  (x,  y)  by  (x,  —  y).     Hence  the  locus  is  symmetrical 
with  respect  to  the  axis  ofx. 

The  curve  is  called  a  parabola. 

3.  Draw  the  locus  of  the  equation 


(8) 


4,y  =  x^. 


X 

.  y 

X 

y 

0 

0 

0 

0 

1 

i 

-1 

-i 

n 

II 

-n 

-w 

2 

2 

-2 

-2 

2| 

3|| 

-^ 

-m 

3 

6f 

-3 

-Of 

H 

lOil 

-H 

-  lOil 

Solution.    First  step.   Solv- 
ing for  y, 


y- 


■■\x\ 


For  example,  if 


Second  step.  Assume  val- 
ues for  X  and  compute  y. 
Values  of  x  must  he  taken 
between  the  integers  in  order 
to  give  points  not  -too  far 
apart. 


^—  ^2"? 


7,  _  1    .    1  2  5  _  1  2  5  _  Q  2  9      (^\o 

Third  step.     Plot  the  points  thus  found. 

Fourth  step.     The  points  determine  the  curve  of  the  figure. 

Discussion.     1.    From   the    given    equation  (8),  x  and  y  in- 


CURVE    AND   EQUATION 


45 


^ 

i 

1 

/ 

f 

\ 

-[ 

1^^ 

1^ 

) 

/ 

1 

/ 

'  J 

J 

r 

X 

r 

) 

/ 

0 

\ 

/ 

/ 

f 

■X 

r/ 

) 

' 

- 

— 

J 

— 

^ 

crease  simultaneously,  and  therefore  the  curve  extends  out  in- 
definitely from  both  axes. 

2.  In  (8)  there  are  no  even  powers  nor 
constant  term,  so  that  by  changing  signs  the 
equation  may  be  written  in  the  form 

K-y)  =  {-xf, 

replacing  {x,  y)  by  {~x,  -y). 

Hence  the  locus  is  symmetrical  with  respect 
to  the  origin. 

The  locus  is  called  a  cubical  parabola. 

20.  Symmetry.  In  the  above  examples 
we  have  assumed  the  definition : 

If  the  points  of  a  curve  can  be  arranged 
in  pairs  which  are  symmetrical  with  respect  to  an  axis  or  a 
point,  then  the  curve  itself  is  said  to  be  symmetrical  with  re- 
spect to  that  axis  or  point. 

The  method  used  for  testing  an  equation  for  symmetry  of  the 
locus  was  as  follows :  if  (x,  y)  can  be  replaced  by  {x,  —  y) 
throughout  the  equation  without  affecting  the  locus,  then  if 
(a,  b)  is  on  the  locus,  (a,  —  b)  is  also  on  the  locus,  and  the 
points  of  the  latter  occur  in  pairs  symmetrical  with  respect  to 
XX',  etc.     Hence 

Theorem  II.  If  the  locus  of  an  equation  is  unaffected  by  re- 
2olacing  y  by  —  y  throughout  its  equatioyi,  the  locus  is  symmetrical 
with  respect  to  the  axis  of  x. 

If  the  locus  is  unaffected  by  changing  x  to  —  x  throughout  its 
equ^-^.ion,  the  locus  is  symmetrical  tvith  respect  to  the  axis  of  y. 

If  the  locus  is  unaffected  by  changing  both  x  and  y  to  —x  and 
—  y  throughout  its  equation,  the  locus  is  symmetrical  tvith  respect 
to  the  origin. 

These  theorems  may  be  made  to  assume  a  somewhat  different 
form  if  the  equation  is  algebraic  in  x  and  y.     The  locus  of  an 


46  ELEMENTARY   ANALYSIS 

tlgebraic  equation  in  the  variables  x  and  y  is  called  an  algebraic 
curve.     Then  from  Theorem  II  follows 

Theorem  III.  Symmetry  of  an  algebraic  curve.  If  no  odd 
powers  of  y  occur  in  an  equation,  the  locus  is  symmetrical  with 
respect  to  XX' ;  if  no  odd  powers  of  x  occur,  the  locus  is  sym- 
metrical with  respect  to  YY\  If  every  term  is  of  even^  degree, 
or  every  term  of  odd  degree,  the  locus  is  symmetrical  with  respect 
to  the  origin. 

21.   Further  discussion.     In  this  section  we  treat  of  three 
more  questions  which  enter  into  the  discussion  of  an  equation. 
Is  the  origin  on  the  curve  ? 
This  question  is  settled  by 

Theorem  IV.  The  locus  of  an  algebraic  equation  passes  through 
the  origin  ivhen  there  is  no  constant  term  in  the  equation. 

Proof  The  coordinates  (0,  0)  satisfy  the  equation  when 
there  is  no  constant  term.  Hence  the  origin  lies  on  the  curve 
(Corollary,  p.  30).  q.e.d. 

What  values  of  x  and  y  are  to  be  excluded  ? 
Since  coordinates  are  real  numbers  we  have  the 

Rule  to  determine  all  values  of  x  and  y  which  must  he  excluded. 

Solve  the  equation  for  x  in  terms  of  y,  and  from  this  result  de- 
termine, all  values  of  y  for  ivhich  the  computed  value  of  x  will  be 
imaginary.     These  values  of  y  must  be  excluded. 

Solve  the  equation  for  y  in  terms  of  x,  and  from  this  result  de- 
termine all  values  of  x  for  which  the  computed  value  of  y  will  be 
im^aginary.     These  values  of  x  must  be  excluded. 

The  intercepts  of  a  curve  on  the  axis  of  x  are  the  abscissas 
of  the  points  of  intersection  of  the  curve  and  XX. 

The  intercepts  of  a  curve  on  the  axis  of  y  are  the  ordinates 
of  the  points  of  intersection  of  the  curve  and  YY^. 

*  The  constant  term  must  be  regarded  as  of  even  (zero)  degree. 


CURVE   AND   EQUATION 


47: 


Rule  to  find  the  intercepts.  ? 

Substitute  y  =  0,  and  solve  for  real  values  of  x.  This  gives  the 
intercepts  on  the  axis  of  x. 

Substitute  x=:Oy  and  solve  for  real  values  ofy.  TJiis  gives  the 
intercepts  on  the  axis  of^. 

The  proof  of  the  rule  follows  at  once  from  the  definitions. 

The  rule  just  given  explains  how  to  answer  the  question : 

What  are  the  intercepts  of  the  locus  ? 

22.  Directions  for  discussing  an  equation.  Given  an  equa- 
tion, the  following  questions  should  be  answered  in  order  before 
plotting  the  locus. 


7s  the  origin  on  the  locus  t  %/y  ^^^JCy<l4. 
Is  the  locus  symmetrical  with  respect  to  the  axes  or  the  origin? 
What  are  the  intercepts  ? 
What  values  of  x  and  y  must  be  excluded  ? 
Is  the  curve  closed  or  does  it  pass  off  indefinitely  far  ? 
Answering    these    questions   constitutes   what   is   called   a 
general  discussion  of  the  given  equation. 

EXAMPLE 

Give  a  general  discussion  of  the  equation 
(1)  aj2-4  2/'  +  16?/  =  0. 

Draw  the  locus. 


1. 
2. 
3. 
4. 

5. 


(O^i) 


48  ELEMENTARY   ANALYSIS 

1.  Since  the  equation  contains  no  constant  term,  the  origin 
is  on  the  curve. 

2.  The  equation  contains  no  odd  powers  of  x ;  hence  the 
locus  is  symmetrical  with  respect  to  YY'. 

3.  Putting  y=Oj  we  find  x  =  0,  the  intercept  on  the  axis  of 
X.  Putting  x  =  0,  we  find  y  =  0  and  4,  the  intercepts  on  the 
axis  of  y. 

4.  Solving  for  x, 


(2)  x  =  ±2Vy'-4.y, 

All  values  of  y  must  be  excluded  which  make  the  expression 
beneath  the  radical  sign  negative.  Now  the  roots  of  y^  —  4:y  =  0 
are  y  =  0  and  y  =  4:.  For  any  value  of  y  between  these  roots, 
y^  —  4:y  is  negative.  For  example^  y=^  gives  4  —  8  =,—  4. 
Hence  all  values  of  y  between  0  and  4  must  be  excluded. 

Solving  for  y, 


(3)  y  =  2±^Vx'-\-W. 

Hence  no  value  of  x  is  excluded,  since  x'^  + 16  is  positive  for 
all  values  of  x. 

5.  From  (3),  y  increases  as  x  increases,  and  the  curve  ex- 
tends out  indefinitely  far  from  both  axes. 

Plotting  the  locus,  using  (2),  the  curve  is  found  to  be  as  in 
the  figure.     The  curve  is  a  hyperbola. 

PROBLEMS 

1.  Grive  a  general  discussion  of  each  of  the  following  equa- 
tions and  draw  the  locus. 


/< 


(a)  af-Ay^O.  /(g)  a^-/  +  4  =  0. 


(c)-i?'  +  4?/2-^6='0..  1/  (i)  xy-i  =  0. 

('0^  ^:  +  f- 18^4  U^Or^  y  +.  *''  =  0- 


(e)  a:^- 4  2/2 -16  =  0.  <f,-(k)  Ax-i^  =  0.  ^^^^ 

(/)  a;2  _  4  /  +  IC  =  0.  ^(0  6  X  -t  =  0.    Y^^^ 


CURVE   AND   EQUATION  49 

(m)  5x-y-j-9f  =  0,  (q)  x^ -{-4.if -^S  y  =  0,' 

(n)  9  y-^  —  x^  =  0.  (r)  x^  -{- 4:  xy  -\- 5  y^  =  4:, 

(o)  9y'-\-x^  =  0,  (s)  x'-i-4.xy  +  y^  =  3. 

(p)  a^  —  y^  -\-6x  =  0. 

2.  Determine  the  general  nature  of  the  locus  in  each  of  the 
■following  equations  by  assuming  particular  values  for  the 
arbitrary  constants,  but  not  special  values,  that  is,  values  which 
give  the  equation  an  added  peculiarity.^ 

(a)  y^  =  2  mx.  (/)  x^  —  y'^=  a?, 

(b)  x^  —  2  my  =  tti^,  (g)  x^  -\-y^  z=  7^. 

(c)  ^  +  -^^  =  1  (^*)  ^^  +  2/'  =  2  rx. 
^'      ^->'         *  (0  x^  +  y'^  =  2ry. 

(d)  2xy  =  a\  (J)  ay''  =  ^. 

(e)  t^vl^l,  Qc)  ahj=:^. 
^^  a'     h' 

3.  Draw  the  locus  of  the  equation 

y-  z=z{x  —  a)(x  —  h)(x—  c), 

(a)  when  a  <b<cc.  (c)  when  a<b,  b  =  c. 

(b)  w^hen  a  =  b<  c.  (d)  when  a  =  b  =  c. 

The  loci  of  the  equations  (a)  to  (/)  in  problem  2  are  all  of 
the  class  known  as  conies,  or  conic  sections, — curves  following 
straight  lines  and  circles  in  the  matter  of  their  simplicity 

A  conic  section  is  the  locus  of  a  point  whose  distances  from  a 
fixed  point  and  a  fixed  line  are  in  a  constant  ratio. 

4.  Show  that  every  conic  is  represented  by  an  equation  of  the 
second  degree  in  x  and  y. 

Hint.  Take  YY'  to  coincide  with  the  fixed  line,  and  draw  XX'  through  the 
fixed  point.    Denote  the  fixed  point  by  (p,  0)  and  the  constant  ratio  by  e. 

AiifS.   (l  —  e'^yx" -\-y'  —  2px -{-p^  =  Q. 

*  For  example,  in  {c()  and  (6)  in  =  0  is  a  special  value.  In  fact,  in  all  these 
examples  zero  is  a  special  value  for  stny  constant. 


50  ELEMENTARY   ANALYSIS 

5.  Discuss  and  plot  the  locus  of  the  equation  of  problem  4, 

(a)  when  e=l.     The  conic  is  now  called  2i  parabola  (see  p.  44). 

(b)  when  e<l.     The  conic  is  now  called  an  ellipse  (see  p.  43). 

(c)  when  e  >1.    The  conic  is  now  called  a  hyperbola  (see  p.  48). 

6.  A  point  moves  so  that  the  sum  of  its  distances  from  the 
two  fixed  points  (3^  0)  and  (—3,  0)  is  constant  and  equal  to  10. 
What  is  the  locus  ?  Ans.  Ellipse  l^x^  +  25?/^  =  400. 

7.  A  point  moves  so  that  the  difference  of  its  distances  from 
the  two  fixed  points  (5,  0)  and  (—  5,  0)  is  constant  and  equal 

,  to  8.     What  is  the  locus  ?         Ans.  Hyperbola  9  o;^— 167/^=144. 

23.  Points  of  intersection.  If  two  curves  whose  equations 
are  given  intersect,  the  coordinates  of  each  point  of  intersection 
must  satisfy  both  equations  when  substituted  in  them  for  the 
variables  (Corollary,  p.  30).  In  Algebra  it  is  shown  that  all 
values  satisfying  two  equations  in  two  unknowns  may  be  found 
by  regarding  these  equations  as  simultaneous  in  the  unknowns 
and  solving.     Hence  the 

Rule  to  find  the  points  of  intersection  of  two  curves  whose  equa- 
tions are  given. 

Consider  the  equations  as  simultaneous  in  the  coordinates,  and 
solve  as  in  Algebra. 

Arrange  the  real  solutions  in  corresponding  pairs.  These  will 
be  the  coordinates  of  all  the  points  of  intersection. 

Notice  that  only  real  solutions  correspond  to  common  points 
of  the  two  curves,  since  coordinates  are  always  real  numbers. 

EXAMPLES 

1.   Find  the  points  of  intersection  of 

(1)  •        x-ly  +  2^  =  0, 

(2)  x'  +  y''=25. 
Solution.     Solving  (1)  for  x, 

(3)  x=^7y-~25. 


CURVE   AND  EQUATION 


51 


"Fa 

^ 

^^ 

\^ 

...^ 

Xi-} 

H 

,3; 

/^ 

\ 

-*" 

\ 

\ 

'■ 

0 

\ 

j 

/ 

\ 

/ 

N 

*^ 

y 

Substituting  in  (2), 

(7  2/ -25/ +  2/2  =  25. 
Reducing, 

,-.  y  =3  and  4. 

Substituting   in  (3)  [iio^ 
in  (2)], 

0^  =  — 4  and  +3. 

Arranging,  the  points  of 
intersection  are  (—4,  3)  and  (o,  4).     A7is, 

In  the  figure  the  straight  line  (1)  is  the  locus  of  equation 
(1),  and  the  circle  the  locus  of  (2). 

2.   Eind  the  points  of  intersection  of  the  loci  of 

(4)  2a^  +  3if  =  S5, 

(5)  3x'-Ay  =  0. 
Solution.     Solving  (5)  for  x^, 

(G)  x'=iy. 

Substituting  in  (4)  and  reducing, 
9f^  +  Sy-105  =  0, 

/.yz=z3  and  —  %^-. 
Substituting  in  (6)  and  solving. 


aj=±2and  ±iV-210. 

Arranging  the  real  values,  we  find  the  points  of  intersection 
are(+2,  3),  (-2,  3).     Ans. 

In  the  figure  the  ellipse  (4)  is  the  locus  of  (4),  and  the  pa- 
rabola (5)  the  locus  of  (o). 


62 


ELEMENTARY   ANALYSIS 


1,. 


PROBLEMS 

Find  the  points  of  intersection  of  the  following  loci. 


jx-lly  +  l  =  0 

x-{-y  —  2  =  0 
Ans.  a,  f). 

x  —  y=:5) 
Ans.  (6,  1). 
■^y  =  Sx  +  2\ 

zyx'+f-=4:  j' 

Ans.  (0,2),  (-f,- 
y^  =  16x  ] 


x'  +  f  =  Al^ 

xy=^20         J  *    " 
Ans.  (±5,  ±4),  f  +  4,±5). 


g    y'  =  22M] 
x'=2py] 


Ans.  (0,  0),  {2p,  2p). 


I)- 


9 


4.x^jry''  =  ^\ 


4. 


y  —  X  =  0] 

Ans.  (0,  0),  (16,  16). 

a;2  +  2/' 


:a" 


5.    "    -       "  .        I 
^x  +  y  +  a  =  0  J 


^n.(0,-a),(-^,^). 


6. 


-/=:16 

Ans.  (±4V2,  4). 


^n8.   (i  2),  (i   -2). 

^2  +  2/'  =  100 
10.     9      9  a; 

^7is.  (8,6),  (8,  -6). 

a?  A- y- z=i  ^  a^  ^ 

11-      2       1  h 

a;^  =  4:  ay         J 

^ns.  (2  a,  a),  (—2  a,  a). 


Find  the  area  of  the  triangles  and  polygons  whose  sides  are 
the  loci  of  the  following  equations. 

■   .12.   3x  +  y  +  4:  =  0,3x-5y  +  3^  =  0,3x-2y  +  l=0. 

Ans.  36. 

13.  x  +  2y  =  5.^2x  +  y  =  7,y  =  x-^l, 

14.  x-\-y  =  a,  x  —  2y  =  Aa,  y  —  x-{-7a  =  0. 

15.  a;  =  0, 2/ =  Oj  a;  =  4,  ?/ =  —  6. 

16.  x  —  y  =  0,x  +  y  =  OyX  —  y  =  a,x  +  y=b. 


Ans.  f . 
Ans.  12  a?. 
Ans.  24. 

Ans.  — 


17.  y=z3x—^,y^3x  +  5,2y  =  x- 


3,  2.?/  =  a; +  14. 

Ans.  5(S. 


CURVE   AND  EQUATION  53 

18.  Find  the  distance  between  the  points  of  intersection  of 
the  curves  3  a^  —  2  ?/  +  6  =  0,  a.-  -f  ?/-  =  9.  Ans.  ||  Vi3. 

19.  Does  the  locus  of  y^  =  4tx  intersect  the  locus  of 
2a^  +  32/  +  2  =  0?  Ans.  Yes.     . 

20.  For  what  value  of  a  will  the  three  lines  ^x-\-y  —  2  =  0^ 
ax-\-2y  —  Z  =  0,  2  a;  —  7/  —  3  =  0  meet  in  a  point  ?  Ans.  a  =  5. 

21.  Find  the  length  of  the  common  chord  of  x'^-\-y^  =  13 
and  y^  =  3x-\-3.  Ans.  6. 

22.  If  the  equations  of  the  sides  of  a  triangle  are  x-\-7y 
+  11  =  0,  3x+y-7  =  0,  x-3y-{-l=:0,  find  the  length  of 
each  of  the  medians.  Aiis.  2  V5,  |V2,  ^VlTO. 


yr\-  y-  ^' 


CHAPTER   IV 

STRAIGHT   LINE   AND   CIRCLE 

24.   The  degree  of  the  equation  of  any  straight  line.     It 

will  now  be 'shown  that  any  straight  line  is  represented  by  an 
equation  of  the  first  degree  in  the  variable  coordinates  x  and  y. 

Theorem.     The  equation  of  the  straight  line  passing  through  a 
point  B(Oy  b)  on  the  axis  of  y  and  having  its  slope  equal  to  m  is 

(I)  y  =  ntiT'  +  &. 

Proof     First  step.     Assume  that  F(x,  y)  is  any  point  on 
the  line. 

Second  step.     The  given  condition  may  be  written 

slope  of  PB  =  m. 

Third  step.     Since  by  (II),  p.  17, 

•       slope  of  P^  =  ^^, 
x  —  0 

[Substituting  (ic,  y)  for  (iCi,  y{)  and  (0,  h)  for  (x2,  2/2)] 
then  ^~    =  m,  or  y  =  mx  +  h.  q.e.d. 

X 

In  equation  (I),  m  and  &  may  have  any  values,  positive, 
negative,  or  zero. 

Equation  (I)  will  represent  any  straight  line  which  intP'^ 
sects  the  ?/-axis.  But  the  equation  of  any  line  parallel  t<» 
?/-axis  has  the  form  x  =  a  constant,  since  the  abscissas  of  all 
points  on  such  a  line  are  equal.  The  two  forms,  y  =  mx  +  b 
and  X  =  constant,  will  therefore  represent  all  lines.  Each  of 
these  equations  being  of  the  first  degree  in  x  and  y,  we  have 

64 


STRAIGHT   LINE   AND   CIRCLE  55 

Theorem.  TJie  equation  of  any  draight  line  is  of  the  first 
degree  in  the  coordinates  x  and  y. 

25.  Locus  of  any  equation  of  the  first  degree.  Tlie  ques- 
tion now  arises :  Given  an  equation  of  the  first  degree  in  the 
coordinates  x  and  y,  is  the  locus  a  straight  line  ? 

Consider,  for  example,  the  equation 

(1)  3i»-2y  +  8  =  0. 

Let  us  solve  this  equation  for  y.     This  gives 

(2)  y  =  ix^L 
Comparing  (2)  with  the  formula  (I), 

y  =  mx  +  b, 
we  see  that  (2)  is  obtained  from  (I)  if  we  set  m  =  f ,  6  =  4. 
Now  in  (I)  m  and  b  may  have  any  values.  The  locus  of  (I)  is, 
for  all  values  of  m  and  6,  a  straight  line.  Hence  (2),  or  (1),  is 
the  equation  of  a  straight  line  through  (0,  4)  with  the  slope 
equal  to  f .  This  discussion  prepares  the  way  for  the  general 
theorem. 

The  equation 

(3)  Ax-hBy-{-C  =  0, 

where  A,  B,  and  C  are  arbitrary  constants,  is  called  the  general 
equation  of  the  first  degree  in  x  and  y  because  every  equation  of 
the  first  degree  may  be  reduced  to  that  form. 
Equation  (3)  represents  all  straight  lines. 

P^or  the  equation  y  =  mx  +  b  may  be  written  mx  —  ?/  +  6  =  0,  which 
is  of  the  form  (3)  if  ^  =  m,  i?  =—  1,  C  =b  ;  and  the  equation  x  =  con- 
stant may  be  written  x  —  constant  =  0,  which  is  of  the  form  (3)  if  ^  =  1, 
^  =  0,  C  =—  constant. 

Theorem.     The  locus  of  the  general  equation  of  the  first  degree 

Ax  +  By+C=^0 
is  a  straight  line. 

Proof     Solving  (3)  for  ?/,  we  obtain 

(4)  2/  =  --^--- 

^  ^  -^  B        B 


56 


ELEMENTARY   ANALYSIS 


Comparison   with  (1)   shows  that  the    locus    of    (4)   is   the 
straight  line  for  which 

G 


m- 


B' 


or 


If,  however,  B  =  0,  the  reasoning  fails. 
But  if  jB  =  0,  (3)  becomes 

Ax +  0  =  0, 

C 


The  locus  of  this  equation  is  a  straight  line  parallel  to  the 
Y-axis.     Hence  in  all  cases  the  locus  of  (3)  is  a  straight  line.  '\ 

Q.E.D. 

Corollary.      TJie  slope  of  the  line 

Ax  +  By-j-C=0 

A 

is  m  = ;  that  is,  the  coefficient  of  x  with  its  sign  changed 

JB 

divided  by  the  coefficient  of  y, 

26.  Plotting  straight  lines.  If  the  line  does  not  pass 
through  the  origin  (constant  term  not  zero,  p.  46),  find  the 
intercepts  (p.  47),  mark  them  off  on  the  axes,  and  draw  the 
line.  If  the  line  passes  through  the  origin,  find  a  second  point 
(p.  37)  whose  coordinates  satisfy  the  equation. 

EXAMPLE 

Plot  the  locus  of3aj  —  2/  +  ^  =  ^-     Find  the  slope. 

Solution.     Letting  y  =  0  and  solving  for  x, 
we  have 

.T  =  —  2=  intercept  on  aj-axis. 
Letting  a?  =  0  and  solving  for  y,  we<have 
2/  =  6  =  intercept  on  ?/-axis. 

X        The  required  line  passes  through  the  points 

(-2,0)  and  (0,  6). 


STRAIGHT   LINE   AND   CIRCLE  57 

To  find  the  slope.  Comparison  with  the  general  equation 
(3)  shows  that  A  =  3,  B  =  -l,  0=6.     Hence  m  =  --^3. 

Otherwise  thus.  Reduce  the  given  equation  to  the  form 
y  =  7nx  -\-bhj  solving  it  for  y.  This  gives  y  =  3  x  -\-6.  Hence 
m  =  3,  6  =  6,  as  before. 

PROBLEMS 

'  1.    Find  the  intercepts  and  the  slope  of  the  following  lines 
and  plot  the  lines. 

(a)  2x  +  3y  =  6.  ^  Ans.  3,  2 ;  77i  =  —  f . 

(b)  x  —  2y-^5  =  0.  Alls.  —5,2^-,  m  =  ^. 

(c)  3x  —  y  +  3  =  0.  Ans.   —1,3;  m  =  3. 

(d)  5x  +  2y-G  =  0.  Ans.  f,  3;  m  =  -f 

2.  Plot  the  following  lines.     Find  the  slope. 

(a)  2x-3y  =  0.  (c)  3  x -{-2  y  =  0. 

(b)  y-4.x  =  0.  (r?)  x-3y  =  0, 

3.  Find  the  equations,  and  reduce  them  to  the  general  form, 
of  the  lines  for  which 


(o.)  m  =  2,b  =  -3. 

Ans. 

2x-y-3  =  0. 

(h)  m  =  -i,&  =  |. 

A71S. 

x-j-2y-3  =  0. 

(c)  m  =  i  Z;  =  -|. 

Ans. 

4.x-10y-25  =  0, 

(d)  «  =  ^,6  =  -2. 

A71S. 

X-y-2  =  0. 

(e)  a^^^,b  =  3. 

Ans. 

x-^y-3  =  0. 

Hint.     Substitute  in  ?/  =  mx  +  h  and  transpose.   ^ 

4.    Select  pairs  of  parallel  and  perpendicular  lines  from  the 
foUowiuQ^. 


ARY  ANALYSIS 


(a) 


(P) 


(P) 


Zi :  y  =  S  X  -  3. 


Arts,  Li  II  Ly,  L2I.  A. 


Ans.  XiXJvg. 


Ans.  L2I.LQ, 

I      7^ 


L^i.y^ -3x-\-2. 
7.3 -4/ =  2.^  +  7. 

'  Li:x  +  3y=z0. 
L2:Sx  +  7j-\-l  =  0. 
.L3:9x-3tj  +  2  =  0. 

L^:2x-5y  =  S, 

L2:5y-\-2x  =  S. 

[Ls:S5x-Uy  =  S. 

5.  Show  that  the  quadrilateral  whose  sides  are  2  aj— 3?/ +4=0, 
3  X  —  y  —  2  =  0,  4,x—  6y  ~9  =  0,  and  6x  —  2y  +  4:  =  0  is  a 
parallelogram. 

6.  Find  the  e'quation  of  the  line  whose  slope  is  —  2  which 
passes  through  the  point  of  intersection  of  y  =  3x-\-4:  and 
y  =z  —  x  +  4:.  Ans.  2aj4-2/  —  4  =  0. 

7.  Write  an  equation  which  will  represent  all  lines  parallel 
to  the  line 

(a)  y=:2x  +  7.  (c)  y-3x-^  =  0. 

Q))  y  =  -x  +  9.  (d)  2y-4:X-\-3  =  0. 

8.  Find  the  equation  of  the  line  parallel  to  2  x  —  3y  =-0 
whose  intercept  on  the  F-axis  is  —  2.      Ans.  2x  —  3y  —  6=ti. 


27.  Point-slope  equation.  If  it  is  required  that  a  straight 
line  shall  pass  through  a  given  point  in  a  given  direction,  the. 
line  is  determined. 

The  following  problem  is  therefore  definite: 

To  find  the  equation  of  the  straight  line  passing  through  a  given 
point  Pi(xi,  2/1)  and  having  a  given  slope  m. 

•    Solution.     Let  P(x,  y)  be  any  other  point  on  the  line.     By 
the  hypothesis, 

slope  PPi  =  m. 


^•'•'  (/ 


(1). 


STRAIGHT   LINE   AND   CIRCLE 

y  —  Vi 


59 
(II,  p.  17) 


Clearing  of  fractions  gives  the  formula 
(II)  y-yi^mipc-a^i). 

28.    Two-point  equation.     A  straight  line  is  determined  by 
two  of  its  points.     Let  us  then  solve  the  problem : 

To  find  the  equation  of  the  line  passing  through  tivo  given 
points  l\{x^,  2/i)j  -^2(^2,  2/2)- 

Solution.     The  slope  of  the  given  line  is 

slope  PiP2  =  ^i^=i!-^ 

Xi        X2 

Let  P  (x,  y)  be  any  other  point  on  the  line  P1P2'     Then 
slope  PPi  =  ^-=^. 

Since  P,  Pj,  and  Pg  are  on  one  line,  slope  PP^  =  slope  PiPg. 
Hence  we  have  the  formula 


(III) 


y-vx^yi-vz^ 

OC  —  OCi       Xi  —  X2* 


TJk 


EXAMPLES 

1.   Find  the  equation  of  the  line  passing  through  Pi  (3,  —2) 
whose  slope  is  —  -\. 

Solution.  Use  the  point- 
slope  equation  (II),  substitut- 
ing x^  =  3,  ?/i  =  -2,  m  =  -  i. 
This  gives 

y-\-2=-l(x-3). 

Clearing  and  reducing, 

x-{-4r.y-i-5  =  0. 


60 


ELEMENTARY   ANALYSIS 


2.  Find  the  equation  of  the  line  through  the  two  points 
Pi  (5,  -l)andP2(2,  -2). 

Solution.     Use   the  two-point   equation   (III),   substituting 

Xi  =  5,   yi  =  -  1,   X2  =  2,   2/2  =  -  2.^ 
This  gives 

^+1^-1  +  2^1 
x-5       5-2       3° 

Clearing  and  reducing, 

x-3y-S  =  0. 

The  answer  should  be  checked.  To  do  this,  we  must  prove  that 
the  coordinates  of  the  given  points  satisfy  the  answer.  Thus 
for  Pi,  substituting  x=5,  y=  —  l,  the  answer  holds.  Similarly 
for  Pg.    The  student  should  supply  checks  for  examples  1  and  3. 

3.  Find  the  equation  of  the  line  through  the  point  Pi  (3,  —  2) 
parallel  to  the  line  Li:2x  —  3y—4: 
=  0. 

Solution.  The  slope  of  the  given 
line  Li  equals  |.  Hence  the  slope  of 
the  required  line  also  equals  f  (The- 
orem, p.  18),  and  it  passes  through 
Pi(3,  —  2).  Using  the  point-slope 
equation  (II),  we  have 

.  y+2=i(x-3),  or  2a;-3?/-12=0. 

4.  Find  the  equation  of  the  line  through  the  point  Pi(—  1, 3) 
perpendicular  to  the  line  Lii  5x  —  2  y  +  3  =  0, 

Solution.  The  slope  of  the  given 
line  Xi  equals  |.  Hence  the  slope  of 
the  required  line  equals  —  |  (Theorem, 

~i ^^'^^ P*   -^^)*     ^i^G®   we   know   a   point   Pi. 

"p  *^J  (  —  1?  ^)  0^^  t^^  lii^^j  we  use  the  point- 

slope  equation  (II),  and  obtain 

2/-3  =  -|(.T-f  1),  or   2x  +  5y-13  =  0, 


STRAIGHT   LINE   AND   CIRCLE  61 

PROBLEMS 

1.  Find  the  equation  of  the  line  satisfying  the  following  con- 
ditions, and  plot  the  lines.     Check  the  answers. 

(a)  Passing  through  (0,  0)  and  (8,  2).  Aris.  x  —  4,y  =  0. 

(b)  Passing  through  (—  1,  1)  and  (—  3,  1).      A71S.  y  —  l  =  0. 

(c)  Passing  through  (—3,  1)  and  slope  =  2. 

Ans.  2x  —  y-{-7  =  0. 

(d)  Having  the  intercepts  *  a  =  S  and  b  =  —2. 

Ans.  2x  —  3y  —  6  =  0, 

(e)  Slope  =  —  3,  intercept  on  X-axis  =  4. 

A71S.  3x  +  y  —  12  =  0. 
(/)  Intercepts  a  =  —  3  and  6  =  —  4. 

A7hs.  4a;  +  32/  +  12  =  0. 
(g)  Passing  through  (2,  3)  and  (-2,-3). 

Ans.  3x-2y  =  0, 
(h)  Passing  through  (3,  4)  and  (—4,-3). 

Ans.  X  —  y  -\- 1  =  0. 
(i)   Passing  through  (2,  3)  and  slope  =  —  2. 

Alts.  2x-{-y  —  7=0. 

2.  Find  the  equation  of  the  line  passing  through  the  origin 
parallel  to  the  line  2x  —  Sy  =  4:.  Ans.  2  x  —  3y  =  0. 

3.  Find  the  equation  of  the  line  passing  through  the  origin 
perpendicular  to  the  line  5x-{-y  —  2  =  0.  Ans.  x  —  5y  =  0. 

4.  Find  the  equation  of  the  line  passing  through  the  point 
(3,  2)  parallel  to  the  line  4.x  — y  — 3  =  0. 

Ans.  4.x  — y  — 10  =  0. 

5.  Find  the  equation  of  the  line  passing  through  the  point 
(3,  0)  perpendicular  to  the  line  2  x  +  y  —  5  =  0. 

Ans.  x  —  2y  —  3  =  0. 

6.  Find   the   equation   of  the  line  whose  intercept  on  the 
y-axis  is  5  which  passes  through  the  point  (6,  3). 

Ans.  x-\-3y  —  15  =  0. 

*  Intercept  on  ai-axis  =  a,  intercept  on  ?/-axis  =  b.    The  given  points  are 
(3,0)  and  (0,-2). 


62  ELEMENTARY  ANALYSIS* 

7.  Find  the  equation  of  the  line  whose  intercept  on  the 
X-axis  is  3  which  is  parallel  to  the  line  ic  —  47/4-2  =  0. 

Ans.  x  —  4:y  —  S  =  0, 

8.  Find  the  equation  of  the  line  passing  through  the  origin 
and  through  the  intersection  of  the  lines  x  —  2  y-\-3  =  0  and 
x-\-2 y  —  9  =  0.  Ans.  x  —  y  —  0, 

9.  Find  the  equations  of  the  sides  of  the  triangle  whose 
.^vertices  are  (—3,  2),  (3,  —2),  and  (0,  —1). 

Ans.  2x  +  3y  =  0,  x-\-3y  +  3  =  0,  and  x  +  y-\-l  =  0. 

10.  Find  the  equations  of  the  medians  of  the  triangle  in 
problem  9,  and  show  that  they  meet  in  a  point. 

Ans.  x  =  0,7x  +  9y-{-3  =  0,  and  5x  +  9y  +  3  =  0. 

iHint.     To' show  that  three  lines  meet  in  a  pointy  find  the  point  of  inter- 
section of  two  of  them  and  prove  that  it  lies  on  the  third. 

11.  Determine  whether  or  not  the  following  sets  of  points 
lie  on  a  straight  line. 

(a)  (0,  0),  (1,  1),  (7,  7).  Ans.  Yes. 

{h)    (2,  3),  (-4,  -6),  (8,  12).  Ans.  Yes. 

(c)  (3,  4),  (1,  2),  (5,  1).  A71S.  No. 

(d)  (3,  - 1),  (-  6,  2),  (-  f,  1).  Ans.  No. 

(e)  (5,6),  (1,1),  (-1,-1).  Ans.  Yes. 
(/)  (7,  6),  (2,  1),  (6,  -2).  Ans.  No. 

12.  Find  the  equations  of  the  lines  joining  the  middle 
points  of  the  sides  of  the  triangle  in  problem  9,  and  show  that 
they  are  parallel  to  the  sides. 

Ans.  4:X-\-6y  +  3  =  0,  x-{-3y  =  0,  and  x-i-y  =  0. 

13.  Find  the  equation  of  the  line  passing  through  the  origin 
and  through  the  intersection  of  the  lines  x-\-2y  =  l  and 
2x  —  4:y  —  3  =  0.  Ans.  x  +  10y  =  0. 

14.  Show  that  the  diagonals  of  a  square  are  perpendicular. 
Hint.    Take  two  sides  for  the  axes  and  let  the  length  of  a  side  be  a. 


STRAIGHT   LINE   AND   CIRCLE  63 

15.  Show  that  the  line  joining  the  middle  points  of  two 
sides  of  a  triangle  is  parallel  to  the  third. 

Hint.    Choose  the  axes  so  that  the  vertices  are  (0,  0),  (a,  0),  and  (6,  c). 

16.  Two  sides  of  a  parallelogram  are  2x-\-^y  —1  =  0  and 
x  —  ^y -\-4:=:0,  Find  the  other  two  sides  if  one  vertex  is  the 
point  (3,  2).  Ans.  2a;  + 32/-12  =  0  and  cc- 32/  +  3  =  0. 

17.  Find  the  equations  of  the  lines  drawn  through  the  vei^- 
tices  of  the  triangle  whose  vertices  are  (—3,  2),  (3,  —2),  and 
(0,  —  1),  which  are  parallel  to  the  opposite  sides. 

Ans,  The  sides  of  the  triangle  are 

2x  +  ^y  =  0,x-\-^y  +  '6  =  0,x  +  y  +  l=z0, 
The  required  equations  are 

2a;  +  32/  +  3  =  0,  i»  +  32/-3  =  0,  a;  +  3/-l  =  0. 

18.  Find  the  equations  of  the  lines  drawn  through  the  ver- 
tices of  the  triangle  in  problem  17  which  are  perpendicular  to 
the  opposite  sides,  and  show  that  they  meet  in  a  point. 

Ans.  3i»-22/-2  =  0,  3aj-?/  +  ll==0,  a;-2/-5  =  0. 

19.  Find  the  equations  of  the  perpendicular  bisectors  of  the 
sides  of  the  triangle  in  problem  17,  and  show  that  they  meet 
in  a  point.       Ans.  3x  —  2y  =  0,  3x  —  y—  6  =  0,  x  —  y  -{-2  =  0. 

20.  The  equations  of  two  sides  of  a  parallelogram  are 
Sx  —  4:y-{-6  =  0  and  x-{-5y  —  10=  0.  Find  the  equations  of 
the  other  two  sides  if  one  vertex  is  the  point  (4,  9). 

Ans.  Sx—4:y-\-24t  =  0  and  x-{-5y —  4:9  =0, 

21.  The  vertices  of  a  triangle  are  (2, 1),  (—  2,  3),  and  (4,  —  1). 
Find  the  equations  of  (a)  the  sides  of  the  triangle,  (b)  the 
perpendicular  bisectors  of  the  sides,  and  (c)  the  lines  drawn 
through  the  vertices  perpendicular  to  the  opposite  sides. 
Check  the  results  by  showing  that  the  lines  in  (b)  and  (c) 
meet  in  a  point. 

29.   The  angle  which  a  line  makes  with  a  second  line. 

The  angle  between  two  directed  lines  has  been  defined  (p.  16) 


64 


ELEMENTARY   ANALYSIS 


as  the  angle  between  their  positive  directions.  When  a  line  is 
given  by  means  of  its  equation,  no  positive  direction  along  the 
line  is  fixed.  In  order  to  distinguish  be- 
tween the  two  pairs  of  equal  angles  which 
two  intersecting  lines  make  with  each 
other,  we  define  the  angle  which  a  line 
makes  with  a  second  line  to  be  the  positive 
angle  (p.  2)  from  the  second  line  to  the 
first  line. 

Thus  the  angle  which  L^  makes  with 
L2  is  the  angle  0.  We  speak  always  of  the  "  angle  which  one 
line  makes  with  a  second  line/'  and  the  use  of  the  phrase  "  the 
angle  between  two  lines ''  should  be  avoided  if  those  lines  are 
not  directed  lines. 

Theorem.    If  m^  and  mg  ai^e  the  slopes  of  two  lines,  then  the 
angle  0  ivhich  the  first  line  makes  with  the  second  is  given  by 


(IV) 


tan  6 


mi  —  ni2 


1  +  inini2 

Proof  Let  ^i  and  «o  be  the  inclinations  of  Li  and  L2  respec- 
tively. Then,  since  the  exterior  angle  of  a  triangle  equals  the 
sum  of  the  two  opposiffe  interior  angles,  we  have 

In  Fig.  1,  a^  =  0-\-  a^,  or  ^  =  ^i  —  ^2, 


In  Fig.  2, 


a^  =  TT  —  ^  +  Wi,   or  ^  =  TT  +  {a^  —  cii) . 


And  since  (30,  p.  2) 

tan  (tt  +  <^)  =  tan  <^y 


STRAIGHT  LINE  AND  CIRCLE 


65 


ihave,  in  either  case, 


tan  0  - 


■-  tan  {cci  —  a^ 

tan  a^  —  tan  a^ 
1  +  tan  «i  tan  a2 


(by  38,  p.  3) 


But  tan  cci  is  the  slope  of  Li,  and  tan  Wgis  the  slope  of  L2] 
hence,  writing  tan  Ui  =  m^  tan  ^3  =  ^^2?  we  have  (IV). 

In  applying  (IV),  we  remember  that  mg  =  slope  of  the  line 
from  which  0  is  measured  in  the  positive  direction. 


EXAMPLES 

1.    Find  the  angles  of  the  triangle  formed  by  the  lines  whose 
equations  are 

L:2x-3y-6  =  0, 

M:6x-y-6  =  0, 

]S[:6x  +  4.y-25  =  0. 

Solution.  To  see  which  angles  formed 
by  the  given  lines  are  the  angles  of  the  tri- 
angle, we  plot  the  lines,  obtaining  the  tri- 
angle ABC. 

Let  us  find  the  angle  A.  In  the  figure, 
A  is  measured  fi^om  the  line  L.  Hence  in 
(IV),  m2= slope  of  Z/=|,  mi  =  slope  oi  M=.6. 

.'.  tan  A  =  ^  ~^  =  — ,  and  A  =  tan~^  -ff- 
1  +  4     15  ^^ 

Next  find  the  angle  at  B.     In  the  figure,  B  is  measured 

2 

3* 


from  N,     Hence  m2  =  slope  of  A^= 

1  TT 

Hence  m2  = ,  and  B  =  --  - 

m  2 


m^ 


:  slope  of  L- 


Finally,  the  angle  at  C  is  measured  from  the  line  M. 

slope  of  N= 


Hence 


in  (IV)  m2=  slope  of  M=  6,  m^ 
.-,  tan  C 


3. 

2* 


:tan^ii|. 


66 


ELEMENTARY   ANALYSIS 


We  may  verify  these  results.    For  if  jB  =- ,  then  A  =  ~- 
and   hence    (31,   p.    3,    and   2^^   p.   3)    tan  ^  =  cot  (7  = 


C', 


tan  (7' 
which  is  true  for  the  values  found. 

2.    Find  the  equation  of  the  line  through 

(3,  6)  which  makes  an  ang]p,i:>i,.  -  with  the 

o 

line  0?  —  2/  +  6  =  0. 

Solution.  Let  m^  be  the  slope  of  the  re- 
quired line.  Then  its  equation  is  by  (II), 
p.  59, 

(1)  ?/  — 5  =  mi(a;  — 3). 

The  slope  of  the  given  line  is  mg  =  1,  and  since  the  angle 

which  (1)  makes  with  the  given  line  is  - ,  we  have, 

o 

.       TT      mi      ^ 
tan-  =— ^ 


or 


whence 


=  -(2  +  V3). 


1-V3 

Substituting  in  (1),  we  obtain 

2/-5  =  -(2  +  V3)(a^-3), 
or  (2  +  V3)i^  +  ij  -  (11  +  3  V3)  =  0. 

In  Plane  Geometry  there  would  be  two  solutions  of  this 
problem, — the  line  just  obtained  and  the  dotted  line  of  the 
figure.     Why  must  the  latter  be  excluded  here  ? 

In  working  out  the  following  problems,  the  student  should 
first  draw  the  figure  and  mark  by  an  arc  the  angle  desired, 
remembering  that  this  angle  is  measured  from  the  second  line 
to  the  first  in  the  counter-clockwise  direction. 


STRAIGHT   LINE   AND   CIRCLE  67 

PROBLEMS 

1.  Find  the  angle  which  the  line  3x  —  y  +  2  —  0  makes  with 
2x-\-y  —  2  =  0;  also  the  angle  which  the  second  line  makes 
with  the  first,  and  show  that  these   angles   are   siipplemeivf^W 

4      4 

2.  Find  the  angle  which  the  line 

(a)  2  X  — ^iif  +  1  =  0  makes  with  the  line  x  —  2y  +  3  =  0. 

(b)  a;  +  2/  +  l  =  ^  i^^^^s  with  the  line  i»—?/  + 1  =  0.      * 

^     (c)   Sx  —  4:y  +  2  =  0  makes  with  the  line  x-[-3y  —  7  =  0. 

(d)  6x—3y  -{-3  =  0  makes  with  the  line  x  =  6,      *» 

(e)  x  —  7y  +  l  =  0  makes  with  the  line  x-\-2y  —  A  =  0. 

In  each  case  plot  the  lines  and  mark  the  angle  found  by  a 
small  arc. 

Ans.  (a)  tan-'(-J^);  (b)  |;  (c)  tan-H-i/);  (d)  tan-'(-l); 
(e)  Un-\j%). 

3.  Find  the  angles  of  the-fTiangle  whose  sides  are  x  +  3y 
-4  =  0,  3a;-22/  +  l  =  0,  and  ^•-?/4-3  =  0. 

Ans.   tan-i(--y-),  tan-\i),  tan-i(2). 

Hint.  Plot  the  triangle  to  see  which  angles  formed  by  the  given  lines 
are  the  angles  of  the  triangle. 

4.  Find  the  exterior  angles  of  the  triangle  formed  by  the 
lines  5x  — y-\- 3  =  0,  yd-z  2,  X- 4:  y-\- 3  =  0. 

Ans.   tan-i (5),  tan-^(-i),  tan~^  (--!/). 

5.  Find  one  exterior  angle  and  the  two  opposite  interior 
angles  of  the  triangle  formed  by  the  lines  2a;  —  3^— 6  =  0, 
3x  +  4:y  —  12  =  0,x  —  3y-{-6  =  0.  Verify  the  results  by  for- 
mula 37,  p.  3. 

6.  Find  the  angles  of  the  triangle  formed  by  3x-\-2y—4^  =  0, 
x  —  3y-{-(j  =  0,  and  4:X  —  3y  —  10  =  0.  Verify  the  results  by 
the  formula 

tan  A  +  tan  B  +  tan  C  =  tan  ^  tan  B  tan  C,iiA-{-B-{-  (7=  180°. 


Q8  ELEMENTAKY  ANALYSIS 

7.   Find  the  lin^passing  through  the  given  point  and  making 
the  given  angle  with  the  given  line. 

(a)  (2,1),  '~,2x-3y  +  2=:0.  Ans.  5x-y-9  =  0. 

(b)  (l,-3),^,x  +  2y  +  4:=p0.        Ans.  3x  +  y=:^0. 

(c)  (xi,  2/1),  <l>,y  =  mx  +  b.      Ans.  y-y,=  »^  +  tan<^  .^_ 

1  —  m  tan  ^ 

(d)  («!,  Vi),  <i>,Ax+By+G=0. 

30.   Equation   of  the   circle.     Ev^ry  circle  is  determined 
when  its  center  and  radius  are  known. 

Theorem.      TJie  equation  of  the  circle  wJiose  center  is  a  given 
point  (a,  P)  and  wJiose  radius  equals  r  is 

(V)  (oo-ay-\-(y-P)^  =  r^ 

Proof.     First  step.     Assume  that  P(x,  y)  is  any  point  on  the 
locus. 

Second  step.     If  the  center  (a,  /?)  be  denoted  by  C,  the  given 
condition  is 

Third  step.     By  (I),  p.  13, 


Squaring,  we  have  (V).  q.e.d. 

Corollary.      The  equation  of  the  circle  whose  center  is  the  origin 
(0,  0)  and  whose  radius  is  r  is 

If  (Y)  is  expanded  and  transposed,  we  obtain 
( 1 )  x^  +  y''  -2  ax  -2  ^y  +  a^  +  ^^  -  7^  =  0, 


STRAIGHT   LINE   AND   CIRCLE  69 

# 

From  the  form  of  this  equation  we  observe : 

Any  circle  is  defined  by  an  equation  of  the  second  degree  in 
the  variables  x  and  y,  in  which  the  terms  of  the  second  degree 
consist  of  the  sum  of  the  squares  of  x  and  y. 

Equation  (1)  is  of  the  form 

(2)  x'^y'  +  Dx  +  Ey  +  F=0, 
where 

(3)  D=-2a,E=-2p,2^ndiF=a?^P''-r'. 

Can  we  infer,  conversely,  that  the  locus  of  every  equation 
of  the  form  (2)  is  a  circle?  By  adding  \D'  +  \h  to  both 
members,  (2)  becomes 

(4)  (x  +  \Df-^r{y  +  \E)'  =  \{D'^-E'^-4.F), 
In  (4)  we  distinguish  three  cases : 

If  D^  +  ^^  —  4  i^  is  positive,  (4)  is  in  the  form  (V),  and  hence 
the  locus  of  (2)  is  a  circle  whose  center  is  (— |-Z),  —^E)  and 
whose  radius  is  r  =  |:  V-D^  -{-E^  —  4:  F. 

If  D^  +  E^  —  4:F=0,  the  only  real  values  satisfying  (4)  are 
x=  —^^D,  y=—-^E  (footnote,  p.  35).  The  locus,  therefore, 
is  the  single  point  (— |-Z>,  —^E),  In  this  case  the  locus  of 
(2)  is  often  called  a  point  circle,  or  a  circle  whose  radius  is  zero. 

If  i>^  +  -E-  —  4  i^  is  negative,  no  real  values  satisfy  (4),  and 
hence  (2)  has  no  locus. 

The  expression  D^-\-E^—4:F  is  called  the  discriminant  of 
(2),  and  is  denoted  by  ®.     The  result  is  given  by  the 

Theorem.      TJie  locus  of  the  equation 

(VI)  00^  i-  y^  -\-  Doc  +  Ey  +  F  =  0, 

whose    discriminant    is    ^^^-=  ^'^  -\   f?    ■  ^  ^i    '^    determined    as 
follows : 

(a)  When  ®  is  positive  the  locus  is  the  circle  ivnose  center  is 
(— I  D,  —\  E)  and  whose  radius  is  r = | Vi^ + ^^ — 4~^  I V®L 

(b)  When  ©  is  zero  the  locus  is  the  point  circle  (— |-Z>,  —^E), 

(c)  When  ©  is  negative  there  is  no  locus. 


70 


ELEMENTARY  ANALYSIS 


Corollary.      When  E  =  0  the  center  of  (YI)  is  on  the  X-axis, 
and  when  D  =  0  the  center  is  on  the  Y-axis. 

Whenever  in  what  follows  it  is  said  that  (YI)  is  the  equar 
tion  of  a  circle  it  is  assumed  that  ©  is  positive. 


EXAMPLE 

Find  the  locus  of  the  equation  a^^  +  2/^  —  4a;  +  8?/  —  5  =  0. 
Solution.     The  given  equation  is  of  the  form  (YI),  where 

D=-4,  E=:S,F=-5, 
and  hence 


r>K 


®=16jr 64+^  =  100  >0. 

The  locus  is  therefore  a  circle 
whose  center  is  the  point 
(2,  —4)    and   whose   radius   is 

The  equation  Ax^  +  Bxy  + 
Cy'-  +  Dx  -{-Ey  +  F=  0  is  called 
the  general  equation  of  the  second 
degree  in  x  and  y  because  it 
contains  all  possible  terms  in  x  and  y  of  the  second  and  lower 
degrees.  This  equation  can  be  reduced  to  the  form  (YI)  when 
and  only  when  A=0  and  B  =  0.  Hence  the  locus  of  an  equa- 
tion of  the  second  degree  is  a  circle  only  when  the  coefficients 
of  x^  and  y^  are  equal  and  th6  xy-teim  is  lacking. 


31.  Circles  determined  by  three  conditions.  The  equation 
of  any  circle  may  be  written  in  either  one  of  the  forms 

(x-ay  +  (y-(3y=r% 

or  '       x^  +  y'  +  Dx  +  Ey-\-F=0. 

Each  of  these  equations  contains  three  arbitrary  constants. 
To  determine  these  constants  three  equations  are  necessary, 
and  as  any  equation  between,  the  constants   means  that  the 


STRAIGHT   LINE    AND   CIRCLE 


71 


circle  satisfies  some   geometrical  condition,  it  follows  that  a 
circle  may  be  determined  to  satisfy  three  conditions. 

Rule  to  determine  the  equation  of  a  circle  satisfying  three 
conditions. 

First  step.     Let  the  required  equation  he 

(1)  {x~af  +  (jj-lif  =  T% 
or 

(2)  x'  +  f-  +  Dx -\-  Ey  +  r={}, 
as  may  be  more  coyivenient. 

Second  step.  Find  three  equations  between  the  constaiits  a,  p, 
and  r  \_or  D,  E,  and  F^  ivhich  express  that  the  circle  (1)  [or  (2)] 
satisfies  the  three  given  conditions. 

Third  step.  Solve  the  equations  found  in  the  second  step  for 
a,  f3,  and  r  [or  D,  E^  and  F']. 

Fourth  step.  Substitute  the  results  of  the  third  step  in  (1)  [or 
(2)].     Tlie  result  is  the  required  equation. 

EXAMPLES 

1.  Find  the  equation  of  the  circle  passing  through  the  three 
points  Pi(0,  1),  P,(0,  6),  and  P^{^,  0). 

First  solution.  First  step.  Let  the 
required  equation  be 

(3)  x'  +  y^-{-Dx  +  Ey-{-F=0. 

Second  step.  Since  P^,  Po,  and  Pg 
lie  on  (3),  their  coordinates  must  sat- 
isfy (3).     Hence  we  have 

(4)  1  +  ^4-P-O, 

(5)  36  +  6J5;  +  P=0, 
and 

(6)  9  +  3i)+P=0. 


72  ELEMENTARY  ANALYSIS 

Tliird  step.     Solving  (4),  (5),  and  (6),  we  obtain 

^=-7,  i^=6,  i)=-5. 
Fourth  step.     Substituting  in  (3),  the  required  equation  is 

The  center  is  (f ,  ^)  and  the  radius  is  f  V2  =  3.5. 

Second  solution.  A  second  method  which  follows  the  geo- 
metrical construction  for  the  circumscribed  circle  is  the  fol- 
lowing. Find  the  equations  of  the  perpendicular  bisectors  of 
P1P2  ^^d  P1P3.  The  point  of  intersection  is  the  center.  Then 
find  the  radius  by  the  length  formula. 

2.  Find  the  equation  of  the  circle  passing  through  the  points 
Pi(0,  —3)  and  P2(4,  0)  which  has  its  center 
on  the  line  a;  H^r  2  i/  sc  0. 

First  solution.     First  step.     Let  the  re- 
i^x    quired  equation  be 

(7)         x'  +  f  +  Dx  +  Ey  +  F=0. 

Second  step.     Since  P^  and  Pg  lie  on  the 
locus  of  (7),  we  have 

9-3J5;  +  i^=0 
16+4i>  +  P=0. 


(8) 
and 

(9) 


The  center  of  (7)  is  (  —  ~, 


D       E 


line, 


-f+^ 


2 
E 


I,  and  since  it  lies  on  the  given 
=  0,  . 


or 

(10)  D^2E  =  0. 

Tliird  step.     Solving  (8),  (9),  and  (10),  we  obtain 

i>  =  --VS^  =  |,andP=--V-. 


STRAIGHT  LINE  AND   CIRCLE  73 

Fourth  step.  Substituting  in  (7),  we  obtain  the  required 
equation, 

or  5a^  +  52/'-14.T  +  72/-24  =  0. 

The  center  is  the  point  (|,  —  ^-^),  and  the  radius  is  |- V29. 

Second  solution.  A  second  solution  is  suggested  by  Geometry, 
as  follows : 

Find  the  equation  of  the  perpendicular  bisector  of  P^  Pg- 
The  point  of  intersection  of  this  line  and  the  given  line  is  the 
center  of  the  required  circle.  The  radius  is  then  found  by  the 
length  formula. 

PROBLEMS 

1.    Find  the  equation  of  the  circle  whose  center  is 

(a)  (0, 1) and  whose  radius  is  3.        Ans.  x^ +  y^— 2y —  8  =  0. 

(b)  ( — 2, 0)  and  whose  radius  is  2.       Ans.  x^ -\-y^-{-4:X  =  0. 

(c)  ( —3, 4)  and  whose  radius  is  5.       Ans.  x^  -{-  y'^  +  6  x  —  S  y  =  0. 

(c?)  (a,  0)  and  whose  radius  is  a.        Ans.  xr-\-y^  —  2ax  —  0.  \ 

(e)  (0,  /3)  and  whose  radius  is  {3.       Ans.  x'-^  -\-y^  —  2/3x  =  0.   'A  ^// 


,.0.  li,  fj 


(/)  (0,  —  ^)  and  whose  radius  is  ^.      -  Ayis.  x^  -{-y^  +  2  px  =  0.    /^   tT 

2.  Find  the  locus  of  the  following  equations. 

(a)aP  +  y^~6x-16  =  0.  (f)  x^  +  y'^  -  6  x  +4.y  -^  5  =  0. 

(b)3a^  +  3  y'- 10  x-24.y  =  0.   (g)  (x  +  If  +  (y  -  2)^  =  0. 
(c)  x'  +  y'  =  0.  (Ji'^jx'-{-7y'^-4.x-y  =  ^. 

(d)x'  +  y^-Sx-6y-{-25  =  0.    (i)  x'+y'-+2ax-^2by+a^-{-b^=:0. 
(e)  af-\-y'^-2x  +  2y  +  5  =  0.    (j)  or  +  y^  +  i6x  -\-100  =  0. 

3.  Find  the  equation  of  the  circle  which 

(a)  has  the  center  (2,  3)  and  passes  through  (3,  —  2) . 

Ans.   x^  +  y'- -  4.  x^6y -13  =  0. 

(b)  passes  through  the  points  (0,  0),  (8,  0),  (0,  —  6). 

A71S.  x^-}-y'--Sx+6y  =  0. 

(c)  passes  through  the  points  (4,  0),  (—2,  5),  (0,  —3). 

Ans.  19  x^  +  19  ?/'  -f  2  x  -  47  y  -^312  =  0. 


74  ELEMENTARY   ANALYSIS 

(d)  passes  through  the  points  (3,  5)  and  X~^y  '^)  ^^^  ^^s 
its  center  on  the  X-axis.  Ans.  x^  -{-  y^  -\-  4:  x  —  4:6  =  0. 

(e)  passes  through  the  points  (4,  2)  and  (—  6,  —2)  and  has 
its  center  on  the  F-axis.  A7is.  x^  -{-  y^  -{-  5  y  —  30  =0. 

(/)  passes  through  the  points  (5,  —  3)  and  (0,  6)  and  has  its 
center  on  the  line  2x~3y  —  6  =  0. 

Ans.  3i»2  +  32/2_ll4aj_  642/ +276  =  0. 
(g)  h^  the  center  (—  1,  —  5)  and  is  tangent  to  the  J^axis. 

Ans.  x'4-y^-\.2x  +  10y+l=0. 
(h)  passes  through    (1,  0)  and  (5,  0)  and  is  tangent  to  the 
IF-axis.  Ans.  x^  +  y^  —  6x±2-\/5y-\-5=0. 

(i)  passes  through  (0/1),  (5, 1),  (2,  -3). 

Ans.  2x^  +  2y^-10x  +  y-3  =  0. 
(j)  has  the  line  joining  (3,  2)  and  (—  7,  4)  as  a  diameter.       ^ . 

Ans:  x^-{-y^  +  4:X-6y-13  =  0\ 
(Tc)  has  the  line  joining  (3,  —  4)  and  (2,  —  5)  as  a  diameter. 

Ans.  x^+y^-5x  +  9y  +  26  =  0. 
(/)  which  circumscribes   the  triangle   formed   by  a;— 6  =  0? 
X  -\-2y  =  0,  and  x~2y  =  S. 

Ans.  2x^  +  2y'^-21x  +  Sy-^60=0. 
The  following  problems  illustrate  cases  in  which  the  locus 
problem  is  completely  solved  by  analytic  methods,  since  the  loci 
may  be  easily  drawn  and  their  nature  determined. 

LOCUS   PROBLEMS 

1.  Find  the  equation  of  the  locus  of  a  point  whose  distances 
from  the  axes  XX'  and  YY'  are  in  a  constant  ratio  equal  to  |. 

Ans.  The  straight  line  2x  —3y  =  0. 

2.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  whose, 
distances  from  the  axes  of  coordinates  is  always  equal  to  10. 

Ans.  The  straight  line  x-^y  —  10  =  0. 

3.  A  point  moves  so  that  the  difference  of  the  squares  of  its 
distances  from  (3,  0)  and  (0,  —  2)  is  always  equal  to  8.  Find 
the  equation  of  the  locus,  and  plot. 

Ayis.  The  parallel  straight  lines  6a; +4?/ +3  =  0,6a;+42/— 13=0. 


STRAIGHT   LINE   AND   CIRCLE  75 

4.  A  point  moves  so  as  to  be  always  equidistant  from  the 
axes  of  coordinates.     Find  the  equation  of  the  locus,  and  plot. 

A71S.   The  perpendicular  straight  lines  x  -^-y  =  0,x  —  y  =0. 

5.  A  point  moves  so  as  to  be  always  equidistant  from  the 
straight  lines  x  —4z  =  0  and  y  -\-5=  0,  Find  the  equation  of 
the  locus,  and  plot. 

Ans.  The  perpendicular  straight  lines  cc—  y— 9  =  0,x-[-y^l=0. 

6.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  the 
squares  of  whose  distances  from  (3,  0)  and  (—3,0)  always 
equals  68.     Plot  the  locus.  Ans,  The  circle  x^-\-y^  =  25. 

7.  Find  the  equation  of  the  locus  of  a  point  which  moves  so 
that  it^  distances  from  (S,  0)  and  (2,  0)  are  always  in  a  constant 
ratio  equal  to  2.     Plot  the  locus.         Ans,  The  circl^  x^-\-y^=16, 

8.  A  point  moves  so  that  the  ratio  of  its  distances  from 
(2,  1)  and  (  —  4,  2)  is  always  equal  to  2.  Find  the  equation  of 
the  locus,  and  plot.  Ans.  The  circle  3xr-{-3y^—24:X—4:y=0. 

In  the  proofs  of  the  following  theorems  the  choice  of  the  axes 
of  coordinates  is  left  to  the  student,  since  no  mention  is  made 
of  either  coordinates  or  equations  in  the  problem.  In  such 
cases  always  choose  the  axes  in  the  most  convenient  manner 
possible. 

9.  A  point  moves  so  that  the  sum  of  its  distances  from  two 
perpendicular  lines  is  constant.  Show  that  the  locus  is  a  straight 
line. 

Hint.  Choosing  the  axes  of  coordinates  to  coincide  with  the  given 
lines,  the  equation  is  x  +  y  =  constant. 

10.  A  point  moves  so  that  the  difference  of  the  squares  of 
its  distances  from  two  fixed  points  is  constant.  Show  that  the 
locus  is  a  straight  line. 

Hint.  Draw  XX'  through  the  fixed  points,  and  YY'  through  their 
middle  point.  Then  the  fixed  points  may  be  written  (a,  0),  (  —  a,  0),  and 
if  the  "constant  difference"  be  denoted  by  k,  we  find  for  the  locus 
4:ax  =  k  or  4:ax  =—  k. 


716  ^^EEEMWTARY   ANALYSIS 

11.  A  point  moves  so  that  the  sum  of  the  squares  of  its 
distances  from  two  fixed  points  is  constant.  Prove  that  the 
locus  is  a  circle. 

Hint.     Choose  axes  as  in  problem  10. 

12.  A  point  moves  so  that  the  ratio  of  its  distances  from  two 
fixed  points  is  constant.     Determine  the  nature  of  the  locus. 

Ans.  A  circle  if  the  constant  ratio  is  not  equal  to  unity,  and 
a  straight  line  if  it  is. 

13.  A  point  moves  so  that  the  square  of  its  distance  from  a 
fixed  point  is  proportional  to  its  distance  from  a  fixed  line 
through  the  fixed  point.     Show  that  the  locus  is  a  circle. 


CHAPTER   V 

CURVE   PLOTTING 

32.   Asymptotes.     The   following   problems   elucidate  diffi- 
culties arising  frequently  in  drawing  the  locus  of  an  equation. 

EXAMPLES 

1.    Plot  the  locus  of  the  equation 
(1)  xy-2y-4.  =  0. 

Solution.     Solving  for  y^ 


(2) 


y- 


^—  2 
We   observe  at   once,  if  x  =  2, 
This    is    interpreted 


X 

y 

X 

y 

0 

-2 

0 

-2 

1 

-4 

-  1 

-f 

H 

-8 

-2 

-1 

If 

-  16 

-4 

-1 

2 

CO 

-5 

-f 

2i 

16 

^ 

8 

-  10 

-i 

o 

4 

etc. 

etc. 

4 

2 

5 

f 

6 

1 

12 

0.4 

etc. 

etc. 

thus :  The  curve  approaches  the 
line  x^2  as  it  passes  off  to  in- 
finity. The  vertical  line  a;  =  2  is 
called  a  vertical  asymptote. 

In  plotting,  it  is  necessary  to 
assume  values  of  x  differing 
slightly  from  2,  both  less  and 
greater,  as  in  the  table. 

From  (2)  it  appears  that  y  diminishes  and  approaches  zero 
as  X  increases  indefinitely.  The  curve  therefore  extends  in- 
definitely far  to  the  right  and  left,  approaching  constantly 
the  axis  of  x.  The  axis  of  x  is  therefore  a  horizontal  asymptote. 
If  we  solve  (1)  for  a;  and  write  the  result  in  the  form 

4 


77 


y 


78 


ELEMENTARY   ANALYSIS 


it  is  evident  that  x  approaches  2  as  2/  increases  indefinitely. 
Hence  the  locus  extends  both  upward  and  downward  indefi- 
nitely far,  approaching  in  each  case  the  line  x  =  2.  This  curve 
is  called  a  liyperhola. 


n 

1 

\ 

\ 

y 

\ 

^ 

**" 

— ■ 

^ 

— 

— 

"~^ 

uO 

<X) 

p- 

■■ — 

0 

r^ 

,0) 

'r 

"^ 

N 

\ 

^ 

\ 

I 

T 

In  the  problem  just  discussed  it  was  necessary  to  learn  ivhat 
value  X  approached  when  y  became  very  large,  and  also  what 
value  7/  approached  when  x  became  very  large.  These  ques- 
tions, when  important,  are  usually  readily  answered,  as  in  the 
following  example. 


Fm 


y=f 


X 


2.   Plot  the  locus  of  ?/  = ^^  • 

When  X  is  very  great,  v/e  may 
neglect  the  3  in  the  numerator 
(2x  +  3)  and  the  —  4  in  the  denom- 
inator (3  ic  —  4).     That  is,  when  x  is 

very     large,     y  =  --=-        Hence 
ox      o 

2 
y  =  -  is  a  horizontal  asymptote. 


CURVE   PLOTTING  79 

The  equation  shows  directly  that  3a?  —  4  =  0  or  x  =  ^  is  ja 
vertical  asymptote.     Or  we  may  solve  the  equation  for  x  which 

4^  +  3 


gives  X 

Hence,  when  y  is  very  large,  x 


3y-2 

4  ?/  _  4 
''37j~3' 


PROBLEMS 

Plot  each  of  the  following,  and  determine  the  horizontal  and 
vertical  asymptotes. 

1.  (a)  x^  +  y-S  =  0.  (e)  2xy  +  4.x-6y  +  3=0, 

(b)  xy-^x  +  3  =  0,  (/)  y^^2xy-4  =  0. 

(c)  2xy  +  2x  +  3y  =  0.  (g)  xy+x  +  2y  -3=0, 

(d)  x^  +  xy  +  S  =  0. 

2.  (a)  x-y-5  =  0.  (c)  0.^-4.^  +  6  =  0. 
(b)  x'y-y+2x  =  0.                 (d)  a^y-y^S  =  0. 

x^  —  3x  x"^  +  x  2/  —  9 


x' 

-4 

x' 

+  x 

y' 

y- 

-1 

,_.v 

—  2 

(J)y  =  4^-  ««'  =  ^-  {l)12.:-^y 


it-  —  4  y—1  ^  ^  3  —  2/^ 

33.   Natural  logarithms.     The  common  logarithm  of  a  given 
number  ^is  the  exponent  x  of  the  base  10  in  the  equation 

(1)  10"^  =  ^,  or  also  x  =  log^o  ^. 

A  second  system  of  logarithms,  known  as  the  natural  system, 
is  of  fundamental  importance  in  mathematics.  The  base  of 
this  system  is  denoted  by  e,  and  is  called  the  natural  base. 
Numerically  to  three  decimal  places, 

(2)  6  =  2.718. 

The  natural  logarithm  of  a  given  number  N  is  the  exponent  y 
in  the  equation 

(3)  e^  =  N,  or  also  y  =  log,  JSf. 


80  ELEMENTARY   ANALYSIS 

To  find  the  equation  connecting  the  common  and  natural 
logarithms  of  a  given  number,  we  may  take  the  logarithms  of 
both  members  of  (3)  to  the  base  10,  which  gives 

(4)  logio  e^  =  logio  ^,  or  2/ logio  e  =  logio  iV. 

(5)  .  •.  logio  N=  logio  ^  •  logg  N  [using  the  value  of  y  in  (3)]. 

The  equation  shows  that  the  common  logarithm  of  any  num- 
ber equals  the  product  of  the  natural  logarithm  by  the  con- 
stant logio  e.  This  constant  is  called  the  modulus  (=  M)  of  the 
common  system.     That  is 

(6)  Jf  =  logio  e  =  0.434.     Also  —  =  2.302. 

We  may  summarize  in  the  equations, 

X .  V  Common  log  =  natural  log  times  0.434, 

Natural  log  =  common  log  times  2.302. 

Exponential  and  logarithmic  curves.  The  locus  of  the 
equation 

(7)  y  =  e^ 

is  called  an  exponential  curve.  To  compute  values  of  y,  we  use 
logarithms.     Taking  natural  logarithms  of  both  sides, 

(8)  a^  =  log,  2/ =  2.302  logio  2/. 

The  locus  of  (7)  is  therefore  the  curve  whose  abscissas  are 
the  natural  logarithms  of  the  ordinates. 

Discussion.  Since  negative  numbers  and  zero  have  no  loga- 
rithms, y  is  necessarily  positive.  Moreover,  x  increases  as  y 
increases.     The  calculation  must  begin  with  small  values  of  y, 

such  as  ——-.  — — ,  — ,  these  numbers  being  chosen,  since 
1000'  100'  10'  ^  ' 

logio  ~^^  =  logio  ^3  =  logio  10-^  =  -  3,  etc.      (16  and  19,  p.  1) 

The  computation  for  determining  points  on  the  locus  is  set 
down  in  the  table.      We  use  the  Table  of   Art.  2,  p.  4.     If 


CURVE   PLOTTING 


81 


X 

y 

X 

y 

-9.2 

.0001 

0 

1 

-6.9 

.001 

.693 

2 

-4.6 

.01 

1.098 

3 

-2.3 

.1 

,  1.386 

4 

etc. 

etc. 

.  2.302 

10 

2.995 

20 

4.605 

100 

etc. 

etc. 

the  curve  is  carefully  drawn, 
natural  logarithms  may  be 
measured  off.  Thus,  by  meas- 
urement in  the  figure,  if 

2/ =  4.5,   aj=1.55. 

This  discussion  illustrates  the  fact  that 

log,0  =  — 00. 

For  clearly  as  y  approaches  zero,  x  becomes  negatively  larger 
and  larger,  without  limit.  Hence  the  cc-axis  is  a  horizontal 
asymptote. 

More  gener^^lly,  the  locus  of 

(9)  .  y  =  e'^, 

where  A;  is  a  given  constant,  is  an  exponential  curve.  The  dis- 
cussion of  the  difference  of  this  locus  from  the  above  figure 
is  left  to  the  reader. 

The  locus  of  the  equation 

(10)  y  =  \og^^x, 

which  is  called  a  logarithmic  curve,  differs  essentially  from  the 
locus  of  (7)  only  in  its  relation  to  the  axes.  In  fact,  both 
curves  are  exponential  or  logarithmic  curves,  depending  upon 
the  point  of  view. 

The   locus    of   (10)   is   given   in   the    accompanying   figure. 


82 


ELEMENTARY   ANALYSIS 


Clearly,  since  logioO  =  — oo,  the  ?/-axis  is  a  vertical  asymptote. 
The  scales  chosen  are 

unit  length  on  XX'  equals  2  divisions, 
unit  length  on  YY'  equals  4  divisions. 


Compound  interest  curve.  The  problem  of  compound  interest 
introduces  exponential  curves.  For,  if  r  =  rate  per  cent  of 
interest,  n=  number  of  years,  then  the  amount  (=  A)  of  one 
dollar  in  n  years,  if  the  interest  is  compounded  annually,  is 
given  by  the  formula 

A=(l  +  ry. 

For  example,  if  the  rate  is  5  per  cent,  the  formula  is 

(11)  .4  =  (1.05)-". 

If  we  plot  years  as  abscissas  and  the  amount  as  ordinates, 
the  corresponding  curve  will  be  an  exponential  curve.  For, 
by  Art.  2  and  (A), 

log,  1.05  =  2.302  times  .021 

=  .048  (to  three  decimal  places). 

Hence  by  (3),  e^^^  =  1.05,  and  the  equation  (11)  becomes 

(12)  A  =  e-'''-, 

which  is  in  the  form  (9) ;  that  is,  k  =  .048. 


CURVE   PLOTTING 
PROBLEMS 

Di'SLW  *  tlie  loci  of  each  of  the  following. 


1. 

y  =  e-"". 

7. 

y  =  xe-^. 

13. 

2/  =  21ogio^a;. 

2. 

2/  =  e-K 

8. 

s  =  th-\ 

14. 

y  =  logio  Va;. 

3. 

y=e'\ 

9. 

^  =  2e-^ 

15. 

2/=log,(l+e^). 

4. 

y  =  e-'\ 

10. 

y  =  e""'. 

16. 

s=log,o{l  +  2t). 

5. 

y  =  2e-\ 

11. 

y  =  2log^^^x. 

17. 

v  =  log,(l  +  f). 

6. 

y=2e-^\ 

12. 

y  =  loge(l  +  x). 

18. 

X  =  logio  (1  -  y). 

34.  Sine  curves.  As  already  explained  (p.  2),  the  two  com- 
mon methods  of  angular  measurement,  namely,  circular  measure 
and  degree  measure,  employ  as  units  of  measurement  the  radian 
and  the  degree  respectively.  The  relation  between  these  units 
is 

(1)  1  radian  = or  57.29  degrees, 

TT 

1  degree  =  0.0174  radians  or  - — , 
in  which  7r  =  3.14  (or  ^-^-  approximately),  as  usual. 


8        % 

Degrees 

°0 

o            "4 

1       1 

1 

1 1 

IT 

5- 

d 

Radians 

i 

JL 

4 

Equations  (1)  may  be  written 
(2)  TT  radians  =  180  degrees. 

Thus  -  radians  =90°,  -  radians  =45°,  etc.    The  two  scales 
^  4 

laid  off  on  the  same  line  give  the  figure. 

In  advanced  mathematics,  it  is  assumed  that  circular  measure 

is  to  be  used.     Thus  the  numerical  values  of 

*  If  the  shape  only  of  the  curves  1  —  10  is  desired,  we  may  replace  e  hy  the 
approximate  value  3  and  make  the  computation  without  using  logarithms. 


84 


ELEMENTARY   ANALYSIS 


sin  2  X,  X  tan 


TTX 

cos  — 

TTX  6 


4  '      2x 


for  x  =  l,  are  as  follows : 

sm2x=  sin  2  radians  =  sin  114°.59  =  0.909, 

X  tan  — •  =  1  .  tan  [  —  radians  j  =  tan  45"^  =  1, 
4.  [4.  J 


cos  —      COS  (  -  radians 

6  \6  I      cos  30° 


=  0.433. 


2x  2  ~      2 

Let  us  now  draw  the  locus  of  the  equation 

(3)  y  =z  sin  X, 

in  which,  as  just  remarked,  x  is  the  circular  measure  of  an 
angle. 


Solution.  Assuming  values  for  x  and  finding  the  correspond- 
ing number  of  degrees,  we  may  compute  y  by  the  table  of 
Natural  Sines,  Art.  4. 

For  example,  if 

x  =  l,  since  1  radian  =  57°.29, 

y  =  sin  57°.29  =  .843.  [by  (3)] 

In  making  the  calculation  for  plotting,  it  is  convenient  to 
choose  angles  at  intervals  of  say  30°,  and  then  find  x  (in 
radians)  and  y  from  the  Table  of  Art.  4. 


CURVE  PLOTTING 


85 


degrees 

X 

radians 

y 

degrees 

X 

radians 

y 

0 

0 

0 

0 

0 

0 

30 

.52 

.50 

-  30 

-  .52 

-  .50 

60 

1.04 

.86 

-  60 

-1.04 

-  .86 

90 

1.56 

1.00 

-  90 

-  1.56 

-  1.00 

120 

2.08 

.86 

-120 

-2.08 

-  .86 

150 

2.60 

.50 

-150 

-2.60 

-  .50 

180 

3.14 

0 

-180 

-3.14 

0 

Thus   for   30°,    y  =  sin  30°  =  .50.      For    150°,    y  =  sin  150° 
=  sin (180°  -  30°)  =  sin  30°  =  .50  (30,  p.  3). 

The  course  of  the  curve  beyond  B  is  easily  determined  from 
the  relation 

sin  (2  7r-\-x)  =  sin  x. 

Hence  y  =  sin  x  =  sin  (2  tt  +  a?), 

that  is,  the  curve  is  unchanged  if  x-^-^ir  he  substituted  for  x. 
This  means,  however,  that  every  point  is  moved  a  distance  2  tt 
to  the  right.  Hence  the  arc  APO  may  be  moved  parallel  to 
XX'  until  A  falls  on  B,  that  is,  into  the  position  BBC,  and  it 
will  also  be  a  part  of  the  curve  in  its  new  position.  This 
property  is  expressed  by  the  statement :  The  curve  y=^mx 
is  a  periodic  curve  with  a  period  equal  to  2  tt.  Evidently,  the 
curve  crosses  OX  at  intervals  equal  to  a  half  period.  Also, 
the  arc  OQB  may  be  displaced  parallel  to  XX'  imtil  0  falls 
upon  C.  In  this  way  it  is  seen  that  the  entire  locus  consists 
of  an  indefinite  number  of  congruent  arcs,  alternately  above 
and  below  XX'. 

General  discussion.     1.  The  curve  passes  through  the  origin, 
since  (0,  0)  satisfies  the  equation. 

2.    Since  sin(—  a?)  =  —  sin  x,  changing  signs  in  (3), 


or 


—  y  —  —  sm  x^ 

—  y  —  sin(— iK). 


86  ELEMENTARY   ANALYSIS 

Hence  the  locus  is  unchanged  if  (x^  y)  is  replaced  by 
(^—x,  —y)y  and  the  curve  is  symmetrical  with  respect  to  the 
origin  (Theorem  II,  p.  45). 

3.  In  (3),  if  x  =  0, 

y  =  sin  0  =  0  =  intercept  on  the  axis  of  y. 

Solving  (3)  for  x, 


(4) 

X  =  sin~^  y. 

In  (4),  if 

2/  =  0, 

a;  =  sin-^0 

=  riTT,  n  being  any  integer. 

Hence  the  curve  cuts  the  axis  of  x  an  indefinite  number  of 
times  both  on  the  right  and  left  of  0,  these  points  being  at  a 
distance  of  ir  from  one  another. 

4.  In  (3),  X  may  have  any  value,  since  any  number  is  the 
circular  measure  of  an  angle. 

In  (4),  y  may  have  values  from  —  1  to  -}- 1  inclusive,  since 
the  sine  of  an  angle  has  values  only  from  —  1  to  + 1  inclusive. 

5.  The  curve  extends  out  indefinitely  along  XX'  in  both 
directions,  but  is  contained  entirely  between  the  lines  y  =  -\-l, 
y  =  -l. 

The  locus  is  called  the  wave  curve,  from  its  shape,  or  the 
sine  curve,  from  its  equation  (3).  The  maximum  value  of  y  is 
called  the  amplitude. 

Again,  let  us  construct  the  locus  of 

(4)  y  =  2sm'^' 

Solution.  We  now  choose  for  x  the  values  0,  i,  1,  1^,  etc., 
radians,  and  arrange  the  work  as  in  the  table. 

The  figure  represents  a  sine  curve  of  period  6  and  amplitude 
2.  For  the  curve  crosses  the  a:-axis  at  intervals  of  3,  and  the 
maximum  value  of  y  equals  2.  To  draw  any  sine  curve,  after 
the  general  shape  is  known,  it  is  necessary  only  to  find  the 


CURVE  PLOTTING 


87 


X 

radians 

radians 

^TTX 

degrees 

3 

y 

0 

0 

0 

0 

0 

I 

\^ 

30 

.50 

1.00 

1 

\^ 

60 

.86 

1.72 

11 

\^ 

90 

1.00 

2.00 

2 

\^ 

120 

.86 

1.72 

n 

\^ 

150 

.50 

1.00 

3 

W 

180 

0 

0 

amplitude  and  the  period.  The  maximum  values  of  the  ordi- 
nate occur  at  odd  multiples  of  a  quarter  period,  and  the  inter- 
sections with  OX  at  multiples  of  each  half  period. 


PROBLEMS 

Plot  the  loci  of  the  equations  :  ^ 


1. 

^y  =  COS  X. 

5. 

7/  =  cos  1  X. 

8. 

2/  =  3cos^. 
4 

2. 

2/  =  sin  2  X. 

6. 

TTX 
2/  =  COS  —  . 

3. 

2/  ==  cos  2  X. 

4. 

y  =  sin  i  X. 

7. 

2/  =  sin^^\ 

9. 

2/  =  2sin^*. 

*  The  cosine  curve  differs  from  the  sine  curve  only  in  the  position  of  the 
?/-axis.  The  maximum  ordinates  occur  at  multiples  of  half  periods  and  the 
intersections  with  OX  at  odd  multiples  of  quarter  periods. 


88 

10.  ?/  =  3sin^. 

5 

11.  y  =  tsinx, 

12.  ?/  =  taii — . 

13.  2/  =  2  tan  x. 
irx 


ELEMENTARY   ANALYSIS 

20.    y  =  CSC  X. 


15.  v  =  3tan — . 

16.  y  =  cot  oj. 

17.  2/  =  cot  —  . 


21.  y  —  ^QG^x, 

22.  2/ =  CSC  ice. 

23.  ?/  =  sec  — . 


18.    2/ =  4  cot 


Tra; 


24.    '?/  =  csc — . 
^  4 


14.    2/ =  2  tan 

3  19.    ?/  =  sec  X. 

25.  oj  =  sin  2/.     Also  written   y  =  arc   sin  cc   or   sin~^  x,    and 
read,  ^^  the  angle  whose  sine  is  xJ' 

26.  a::  =  2  cos  y,  or  y  =  arc  cos  |^  a;. 

27.  a;  =  tan  y^  or  y  =  arc  tan  cc. 

28.  a:;  =  2  sin  |  Try.  30.    ?/  =  arc  tan  i  a:. 

29.  iK  =  |-  cos  ^  TT?/.  31.    2/  =  ^  ^^^  ^^^  i  ^• 

35.    Addition  of  ordinates.     When  the  equation  of  a  curve 
has.  the  form 

y  =  the  algebraic  sum  of  two  expressions, 

as,  for  example,  y  =  sin  x  +  cos  x,  y  z=^x+  sin^ x,  s  =  6*  +  e""^, 
etc.,  the  principle  known  as  addition  of  ordinates  may  with 
advantage  be  employed.     For  example,  to  construct  the  locus  of 


CURVE   PLOTTING 


89 


(1) 


1       ,   o    •     TTo:^ 


2/2  =  2  sin  ^     (Fig.  6), 


we  employ  the  auxiliary  curves 
(2)  y,  =  ^x    (Fig.  a), 

using  the  same  axis  of  ordinates  but  distinct  axes  of  abscissas. 
Moreover,  the  same  scale  must  be  used  in  both  figures.  The 
ordinates  of  Fig.  b  are  now  added  (in  Fig.  c)  to  the  cor- 
responding ones  in  Fig.  a,  attention  being  given  to  the 
algebraic  signs.    The  derived  curve  A1B1OB2A2  has  the  equation 


(3) 


2/  =  !/i  +  2/2  =  ^^  +  2sin  — 


as  required.     The  locus  winds  back  and  forth  across  the  line 
7j  =  ^x,  crossing  this  line  at  x  =  Oy  ±  4,  ±  S,  ±  12,  etc. 


90  ELEMENTARY   ANALYSIS 

PROBLEMS 

Plot  the  curves : 

1.    ?/  =  1-  cc  +  COS  X.  9.    y  =  e^  —  cos  4  x. 

a?^   ,     .    o  10.    v  =  sin  cc  +  sin  2  x. 

10  .      'JTX    ,  TTX 

11.    ?/=sin hcos— -. 

3.  y  =  sin  0?  +  cos  x.  4  3 

1         o    •    Tra;  12.    v  •=  sin  aa^  +  cos  aoj. 

4.  2/  =  -x  — dsm— .  -^ 


y-- 


4  3  13.    ^  =  2  sin  fl?4-5  cos  0?. 

:^_4cos  — .  14-    2/ =  2  sin  2  a;  H- 3  cos  ice. 

1^  15.    ?/  =  sin  aa?+  sin  bx. 


P^   _J_    /3— «  ^ 

6.    y^^  ^^    ,  16.   2/  =  i^  since. 


2 
7.    y  —  e^  —  sin  2  a?. 


17.  y  =  x  cos  a?. 

18.  ?/ ^  yi^  ce^  sin  cc. 


8.    y  =  ^       ^    .  19.    ?/  =  yi^aj2cos; 


CHAPTER   VI 

FUNCTIONS   AND   GRAPHS 

36.  Functions.  In  many  practical  problems  two  variables 
are  involved  in  such  a  manner  that  the  value  of  one  depends 
upon  the  value  of  the  other.  For  example,  given  a  large  num- 
ber of  letters,  the  postage  and  the  weight  are  variables,  and 
the  amount  of  the  postage  depends  upon  the  weight.  Again, 
the  premium  of  a  life  insurance  policy  depends  upon  the  age 
of  the  applicant.  Many  other  examples  will  occur  to  the 
student. 

This  relation  between  two  variables  is  made  precise  by  the 
definition  : 

A  variable  is  said  to  be  a  function  of  a  second  variable  when 
its  value  depends  upon  the  value  of  the  latter,  and  is  determined 
when  a  definite  value  is  assumed  for  the  latter  variable. 

Thus  the  postage  is  determined  when  a  definite  weight  is 
assumed;  the  premium  is  determined  when  a  definite  age  is 
assumed. 

Consider  another  example : 

Draw  a  circle  of  diameter  5  in.  An  indefinite  number  of 
rectangles  may  be  inscribed  within  this  circle.  But  the  stu- 
dent will  notice  that  the  entire  rectangle  is  determined  as  soon 
as  a  side  is  drawn.  Hence  the  area  of  the  rectangle  is  a  func- 
tion of  its  side. 

Let  us  now  find  the  equation  expressing  the  relation  be- 
tween a  side  and  the  area  of  the  rectangle. 

Draw  any  one  of  the  rectangles  and  denote  the  length  of  its 
base  by  x  in.     Then   by  drawing   a   diagonal   (which   is,    of 

91 


92 


ELEMENTARY   ANALYSIS 


course,  a  diameter  of  the  circle),  the  altitude  is  found  to  be 
equal  to  (2^  —  x^Y.     Hence  if  A  denotes  the  area  in  square 

inclieSy  we  have 

(1)  A  =  x{2b-x')i. 

This  equation  gives  the  functional  \ 
relation  between  the  function  A  and 
the  variable  x.  From  it  we  are  en- 
abled to  calculate  the  value  of  the 
function  A  corresponding  to  any  value 
of  the  variable  x.     For  example : 

if  i»  =  1  in.,  A  =  (24)*  =  4.9  sq.  in. ; 

if  0?  =  3  in.,  A  =  12  sq.  in. ; 

if  i»  =  4  in.,  A  =  12  sq.  in. ;  etc. 

To  obtain  a  representation  of  the  equation  (1)  for  all  values 
of  X,  we  draw  a  graph  of  the  equation.  This  we  do  by  draw- 
ing rectangular  axes  and  plotting 

the  values  of  the  variable  (x)  as  abscissas, 
the  values  of  the  function  (A)  as  ordinates. 

Any  functional  relation  may  be  graphed  in  this  way.     We 
must,    however,    first   discuss    the 
equation  (1). 

The  values  of  x  and  A  are  pos- 
itive from  the  nature  of  the  prob- 
lem. 

The  values  of  x  range  from  zero 
to  5,  inclusive. 

The  student  should  now  choose 
a  suitable  scale  on  each  axis  and 
draw  the  graph.  In  this  case, 
unit  length  on  the  axis  of  abscissas 

represents  1  in.,  and  unit  length  on  the  axis  of  ordinates  rep- 
resents 1  sq.  in.  These  two  unit  lengths  need  not  be  the 
same. 


Inches 


FUNCTIONS   AND   GRAPHS  93 

What  do  im  learn  from  the  graph  ? 

1.  If  carefully  drawn,  we  may  measure  from  the  graph  the 
area  of  the  inscribed  rectangle  corresponding  to  any  side  we 
choose  to  assume. 

2.  There  is  one  horizontal  tangent.  The  ordinate  at  its 
point  of  contact  is  greater  than  any  other  ordinate.  Hence 
this  discovery :  One  of  the  inscribed  rectangles  is  greater  in  area 
than  any  of  the  others,  —  that  is,  there  is  a  maximum  rectangle. 
In  other  words,  the  function  defined  by  equation  (1)  has  a 
maximum  value. 

We  cannot,  of  course,  find  this  value  exactly  by  measure- 
ment.    For  this  purpose  Calculus  is  necessary. 

The  fact  that  a  maximum  rectangle  exists  can  be  seen  in 
advance  by  reasoning  thus :  Let  the  base  x  increase  from  zero 
to  5  in.  The  area  A  will  then  begin  with  the  value  zero  and 
return  to  zero.  Since  A  is  always  positive,  the  graph  must 
have  a  "  highest  point."  Hence  there  is  a  maximum  value  of 
Aj  and  therefore  a  maximum  rectangle. 

Take  one  more  example :  A  wooden  box,  open  at  the  top,  is 
to  be  built  to  contain  108  cu.  ft.  The  base  must  be  square. 
This  is  the  only  condition.  It  is  evident  that  under  this  con- 
dition any  number  of  such  boxes  may  be  built,  and  that  the 
number  of  square  feet  of  lumber  used  will  vary  accordingly. 
If,  however,  we  choose  any  length  for  a  side  of  the  square 
base,  only  one  box  with  this  dimension  can  be  built,  and  the 
material  used  is  determined.  Hence  the  material  used  is  a 
function  of  a  side  of  the  square  base. 

Let  us  now  find  the  functional  relation  between  the  number 
of  square  feet  of  lumber  necessary  and  the  length  of  one  side 
of  the  square  base  measured  in  feet. 

Consider  any  one  box. 

Let  M=^  amount   of  lumber  in 

square  feet ; 

let  cc  =  length     of      side     of 

square  base  in  feet; 


A     / 

/ 

h 


94 


ELEMENTARY   ANALYSIS 


let 
Then 
then 
Hence 


h  =  height  of  the  box  in  feet, 
area  base  =  x^  sq.  ft. ; 


area  sides  =  4  hx  sq.  ft. 
M=  x^  -f-  4  hx. 

But  a  relation  exists  between  h  and  x,  for  the  value  of  M 
must  depend  upon  the  value  of  x  alone,  In  fact,  the  volume 
equals  108  cu.  ft. 

Hence 


hx'  =  10S,   and   h  =  ~' 

x^ 


Therefore 


(2) 


M=:^-}- 


432 


This  equation  enables  us  to  calculate  the  number  of  square 
feet  of  lumber  in  any  box  with  a  given  square  base  which  has 
a  capacity  of  108  cu.  ft.     The  calculation  is  given  in  the  table  : 


X 

M 

0 

CO 

1 

433 

2 

3 

4 
124 

5 

6 

7 

8 

... 

20 

etc. 

feet 

220 

153 

111 

108 

111 

118 

421 

etc. 

sq.  ft. 

Thus,  if 


X 


if 
if 


=  1  ft.,  M=  433  sq.  ft. ; 
x  =  4.  ft.,  M=  124  sq.  ft. ; 
x  =  S  ft.,   M=^  118  sq.  ft. ;  etc. 


The  student  should  now  graph  equation  (2),  choosing  units 
thus: 


FUNCTIONS   AND   GRAPHS  95 

unit  length  on  the  axis  of  abscissas  represents  1  f t. ; 
unit  length  on  the  axis  of  ordinates  represents  1  sq.  ft. 

We  must,  however,  choose  a  very  small  unit  ordinate,  since  the 
values  of  M  are  large. 

A  preliminary  discussion  of  (2)  shows  that  x  may  have  any 
value  (positive). 

What  do  ive  leant  from  the  graph  ? 

(1)  If  carefully  drawn,  we  may  measure  from  the  graph  the 
number  of  square  feet  of  lumber  in  any  box  which  contains 
108  cu.  ft.  and  has  a  square  base. 

(2)  There  is  07ie  horizontal  tangent.  The  ordinate  at  its 
point  of  contact  is  less  than  any  other  ordinate.  Hence  this 
discovery :  One  of  the  boxes  takes  less  lumber  than  any  other ; 
that  is,  M  has  a  minimum  value.  This  point  on  the  graph  can 
be  determined  exactly  by  the  Calculus,  but  careful  measure- 
ment will  in  this  case  give  the  correct  values,  viz.  a?  =  6, 
M— 108.  That  is,  the  construction  will  take  the  least  lumber 
(108  sq.  ft.)  if  the  base  is  6  ft.  square. 

The  fact  that  a  least  value  of  M  must  exist  is  seen  thus. 
Let  the  base  increase  from  a  very  small  square  to  a  very  large 
one.  In  the  former  case  the  height  must  be  very  great,  and 
hence  the  amount  of  lumber  will  be  large.  In  the  latter  case, 
while  the  height  is  small,  the  base  will  take  a  great  deal  of 
lumber.  Hence  Jf  varies  from  a  large  value  to  another  large 
value,  and  the  graph  must  have  a  "lowest  point.'' 

In  the  following  problems  the  student  will  work  out  the 
functional  relation,  draw  the  graph,  and  state  any  conclusions 
to  be  drawn  from  the  figure.  Care  should  be  exercised  in  the 
selection  of  suitable  scales  on  the  axes,  especially  in  the  scale 
adopted  for  plotting  values  of  the  function  (compare  p.  94). 
The  graph  should  be  neither  very  flat  nor  very  steep.  To 
avoid  the  latter  we  may  select  a  large  unit  of  length  for  the 
variable.  The  plot  should  b6  accurate  so  that  the  maximum 
and  minimum  values  of  the  function  may  be  measured. 


96  ELEMENTARY   ANALYSIS 

PROBLEMS 

1.  Eectangles  are  inscribed  in  a  circle  of  radius  r.  Plot  the 
perimeter  P  of  the  rectangles  as  a  function  of  the  breadth  x. 

Ans.  P=2aj  +  2(4r^-a^)J. 

2.  Eight  triangles  are  constructed  on  a  line  of  a  given 
length  h  as  hypotenuse.  Plot  (a)  the  area  A  and  (6)  the 
perimeter  P  as  a  function  of  the  length  x  of  one  leg. 

Ans,  (a)  A  =  ^x (W  -  x^f,     (b)   F=x-\-Ji  +  (h'  -  x'jK 

3.  Eight  cylinders  "^  are  inscribed  in  a  sphere  of  radius  r. 
Plot  as  functions  of  the  altitude  x  of  the  cylinder,  (a)  volume 
V  of  the  cylinder,  (b)  curved  surface  S. 

Ans,  (a)   V=-  (4  t^x-o^),     (b)  /S  =  to (4  r^  -  x^)i, 

4.  Eight  cones  ^  are  inscribed  in  a  sphere  of  radius  r.  Plot 
as  functions  of  the  altitude  x  of  the  cone,  (a)  volume  V  of  the 
cone,  (5)  curved  surface  S, 

Ans,  (a)  V=-(2rx'-:^),     (b)  S  ==7r(4:7^'x' -2  ra^l 
o 

5.  Eight  cylinders  are  inscribed  in  a  given  right  cone.  If 
the  height  of  the  cone  is  7i,  and  the  radius  of  the  base  r,  plot 
(a)  the  volume  V  of  the  cylinder,  (b)  the  curved  surface  S, 
(c)  the  entire  surface  T,  as  functions  of  the  altitude  x  of  the 
cylinder. 

Ans,  (a)   r^'^Qi-xf',  (b)  S  =  ^-Jp  (h-x)-, 

(c)   T=^(h-x){rh+{Ji-r)x), 

6.  Eight  cones  are  circumscribed  about  a  sphere  of  radius  r. 
Plot  as  a  function  of  the  altitude  x  of  the  cylinder,  the  vol- 
ume V  of  the  cone.  ^        t^_  1        ^^^^ 


3     x-2r 

*  Use  formulas  5-9,  p.  1. 


FUNCTIONS   AND   GRAPHS  97 

7.  Right  cones  are  constructed  with  a  given  slant  height  L, 
Plot  as  functions  of  the  altitude  x  of  the  cone,  (a)  the  volume 
V  of  the  cone,  (6)  the  curved  surface  S,  (c)  the  entire  sur- 
face T. 

Ans.  (a)   V=^7r(IJx  —  a:^)  -y 

(b)  jS  =  7rL{L^-x')i. 

8.  A  conical  tent  is  to  be  constructed  of  given  volume  V. 
Plot  the  amount  A  of  canvas  Required  as  a  function  of  the 

radius  x  of  the  base.  Ans.  A  =  \'^  ^  '^ L  . 

X 

9.  A  cylindrical  tin  can  is  to  be  constructed  of  given  vol- 
ume V.     Plot  the  amount  A  of  tin  required  as  a  function  of 

2  V 
the  radius  x  of  the  can.  A7is.  A  =  2  ttx^  -\ 

X 

10.  An  open  box  is  to  be  made  from  a  sheet  of  pasteboard 
12  in.  square,  by  cutting  equal  squares  from  the  four  corners 
and  bending  up  the  sides.  Plot  the  volume  F  as  a  function 
of  the  side  x  of  the  square  cut  out.  Ans.   V=  ct'(12  —  2  a;)l 

11.  The  strength  of  a  rectangular  beam  is  proportional  to 
the  product  of  the  cross  section  by  the  square  of  the  depth. 
Plot  the  strength  /S  as  a  function  of  the  depth  x  for  beams 
which  are  cut  from  a  log  12  in.  in  diameter. 

A71S.    S  =  kx\U4:-x^)^. 

12.  A  rectangular  stockade  is  to  be  built  to  contain  a  cer- 
tain area  A.  A  stone  wall  already  constructed  is  available 
for  one  of  the  sides.  Plot  the  length  L  of  the  wall  to  be  built 
as  a  function  of  the  length  x  of  the  side  of  the  rectangle  paral- 

2  A 
lei  to  the  wall.  Ans.    L  =  -—  -f-  x, 

X 

13.  A  tower  is  100  ft.  high.  Plot  the  angle  y  subtended 
by  the  tower  at  a  point  on  the  ground  as  a  function  of  the  dis- 
tance X  from  the  foot  of  the  tower.  Ans.   y  =  tan~^  — - . 

X 


98  ELEMENTARY   ANALYSIS 

14.  A  tower  50  ft.  high  is  surmounted  by  a  statue  10  ft. 
high.  If  an  observer's  eyes  are  5  ft.  above  the  ground,  plot 
the  angle  y  subtended  by  the  statue  as  a  function  of  the 
observer's  distance  x  from  the  tower. 

Ans.   y  =  tan"^ tan"^  —  • 

15.  A  line  is  drawn  through  a  fixed  point  (a,  b).  Plot  as  a 
function  of  the  intercept  on  XX'  (=x)  of  the  line,  the  area  A 
of  the  triangle  formed  with  the  coordinate  axes. 

A  A  bX^ 

2(x-a) 

16.  A  ship  is  41  mi.  due  north  of  a  second  ship.  The 
first  sails  south  at  the  rate  of  8  mi.  an  hour,  the  second  east 
at  the  rate  of  10  mi.  an  hour.  Plot  their  distance  d  apart 
as  a  function  of  the  time  t  which  has  elapsed  since  they  were 

in  the  position  given.  Ans.    d  =  (164  f  +  656t  +  1681)*. 

17.  Plot  the  distance  e  from  the  point  (4,  0)  to  the  points 
(x,  y)  on  the  parabola  y'^  —  4:X.  Ans,    e=(x^  —  4:X  +  16)^. 

18.  A  gutter  is  to  be  constructed  whose  cross  section  is  a 
broken  line  made  up  of  three  pieces,  each  4  inches  long,  the 
middle  piece  being  horizontal,  and  the  two  sides  being  equally 
inclined.  Plot  the  area  ^  of  a  cross  section  of  the  gutter  as 
a  function  of  the  width  x  of  the  gutter  across  the  top. 

Ans.   A  =  \(x  +  ^)(AS  +  Sx-x')K 

19.  A  Norman  window  consists  of  a  rectangle  surmounted 
by  a  semicircle.     Given  the  perimeter  P,  plot  the  area  ^  as  a 

function  of  the  width  x.  Ans.    A  —  -  xP x^  —  —  x^. 

2  2         8 

20.  A  person  in  a  boat  9  mi.  from  the  nearest  point  of  the 
beach  wishes  to  reach  a  place  15  mi.  from  that  point  along 
the  shore.  He  can  row  at  the  rate  of  4  mi.  an  hour  and 
walk  at  the  rate  of  5  mi.  an  hour.  The  time  it  takes  him 
to  reach  his  destination  depends  on  the  place  at  which  he  lands. 


FUNCTIONS   AND   GRAPHS  99 

Plot  the  time  as  a  function  of  the  distance  x  of  his  landing 
place  from  the  nearest  point  on  the  beach. 

Ans,    Time  =         , 1 


21.  The  illumination  of  a  plane  surface  by  a  luminous  point 
varies  directly  as  the  cosine  of  the  angle  of  incidence,  and  in- 
versely as  the  square  of  the  distance  from  the  surface.  Plot 
the  illumination  /  of  a  point  on  the  floor  10  ft.  from  the  wall, 
as  a  function  of  the  height  iK  of  a  gas  burner  on  the  wall. 

Ans.I= ^ 

(100  + a;-)* 

22.  A  Gothic  window  has  the  shape  of  an  equilateral  tri- 
angle mounted  on  a  rectangle.  The  base  of  the  triangle  is  a 
chord  of  the  window.  The  total  length  of  the  frame  of  the 
window  is  constant.  Express,  plot,  and  discuss  the  area  of 
the  window  as  a  function  of  the  width. 

23.  A  printed  page  is  to  contain  24  sq.  in.  of  printed  matter. 
The  top  and  bottom  margins  are  each  1^  in.,  the  side  margins 
1  in.  each.  Express,  plot,  and  discuss  the  area  of  the  page  as 
a  function  of  the  width. 

24.  A  manufacturer  has  96  sq.  ft.  of  lumber  with  which  to 
make  a  box  with  a  square  base  and  a  top.  Express,  plot,  and 
discuss  the  contents  of  the  box  as  a  function  of  the  side  of  the 
base. 

25.  Isosceles  triangles  of  the  same  perimeter,  12  in.,  are  cut 
out  of  rubber.  Express,  plot,  and  discuss  the  area  as  a  func- 
tion of  the  base. 

26.  Small  cylindrical  boxes  are  made  each  with  a  cover 
whose  breadth  and  height  are  equal.  The  cover  slips  on  tight. 
Each  box  is  to  hold  ir  cu.  in.  Express,  plot,  and  discuss 
the  amount  of  material  used  as  a  function  of  the  length  of  the 
box. 

27.  A  circular  filter  paper  has  a  diameter  of  11  in.  It  is 
folded  into  a  conical  shape.     Express  the  volume  of  the  cone . 


100  ELEMENTAKY  ANALYSIS 

as  a  function  of  the  angle  of  the  sector  folded  over.     Plot  and 
discuss  this  function. 

28.  Two  sources  of  heat  are  at  the  points  A  and  B.  Ee- 
membering  that  the  intensity  of  heat  at  a  point  varies  inversely 
as  the  square  of  the  distance  from  the  source,  express  the  in- 
tensity of  heat  at  any  point  between  A  and  ^  as  a  function  of 
its  distance  from  A.     Plot  and  discuss  this  function. 

29.  A  submarine  telegraph  cable  consists  of  a  central  circu- 
lar part,  called  the  core,  surrounded  by  a  ring.  If  x  denotes 
the  ratio  of  the  radius  of  the  core  to  the  thickness  of  the  ring, 

it  is  known  that  the  speed  of  signaling  varies  as  a?  log  -  •    Plot 

X 

and  discuss  this  function. 

30.  A  wall  10  ft.  high  surrounds  a  square  house  which  is 
15  ft.  from  the  wall.  Express  the  length  of  a  ladder  placed 
without  the  wall,  resting  upon  it  and  just  reaching  the  house, 
as  a  function  of  the  distance  of  the  foot  of  the  ladder  from 
the  wall.     Plot  and  discuss  this  function. 

37.  Notation  of  functions.  Th^  symbol  f(x)  is  used  to 
denote  a  function  of  x,  and  is  read  /  of  x.  In  order  to  distin- 
guish between  different  functions,  the  prefixed  letter  is  changed, 
as  F(x),  <l>(x)yf'{x),  etc. 

During  any  investigation  the  same  functional  symbol  always 
indicates  the  same  law  of  dependence  of  the  function  upon  the 
variable.  In  the  simpler  cases,  this  law  takes  the  form  of  a 
series  of  analytical  operations  upon  that  variable.  Hence,  in 
such  a  case,  the  same  functional  symbol  will  indicate  the  same 
operations  or  series  of  operations,  even  though  applied  to  dif- 
ferent quantities.     Thus,  if 

f(x)  =  x^-9x  +  U, 
then  f(y^=^y^-9y  +  U, 

Also  f(a)  =  a^  —  9  a  + 14, 

/(6  + !)  =  (&  + 1)2 -9(6 +  1)4-14  =  5' -76 +6, 


J 


FUNCTIONS   AND  <>KA?IIS  IQl 

/(0)  =  02_9. 0  +  14  =  14,^.  ,;,.    .      ^ 
/(-l)  =  (-l)^-9(-<l)  +  14  =  24, 
/(3)=32-9.3  +  14  =  -4, 
/(7)=  72  -  9  .  7  + 14  =  0,  etc. 

PROBLEMS 

1.  Given   <!>  (x)  =^  log.o  x.     Find  ct>(2),  <^(1),  <^(5),  ^(a-1), 
<l>(y),<l>(x  +  l),ct>(^x), 

2.  Given    <f>(x)  =  e^.      Find    <^(0),    <^(1),    <^(-l),    <i>(2y), 
<i>{-x). 

3.  Given  fix)  =  sin  2  a:.     Find  /|\  /jY  /(- tt),  /(-  .;), 


/(7r-a;),/(i7r-^),/(f7r  +  5). 
4.    Given  ^  (a;)  =  cos  x.     Prove 


CHAPTER   VII 


DIFFERENTIATION 


38.  Tangent  at  a  point  on  the  graph.     A  glance  at  the 
figure  on  p.  94^  that  is,  the  graph  of  the  equation 


(1) 


X 


in  which  M  represents  the  number  of  square  feet  of  lumber 
required  to  construct  an  open  box  with  a  square  base  to  con- 
tain 108  cu.  ft.,  makes 
clear  the  following  facts. 
When  the  base  is  less 
than  6  ft.  square,  the 
material  decreases  as  the 
size  of  the  base  increases. 
(For  the  graph  is  falling 
to  the  left  of  x=  6.) 
When  the  base  is  more 
than  6  ft.  square,  the 
material  increases  as  the  base  increases.  (For  the  graph  is 
rising  to  the  right  of  a?  =  6.) 

If  we  draw  the  tangent  to  the  graph  at  any  point  to  the  left 
of  a;  =  6,  the  slope  is  clearly  negative,  while  for  any  point  to 
the  rifh4^f  x  =  6,  the  slope  is  positive.  At  the  lowest  point 
x  =  6,  the  slope  is  zero.  Clearly,  then,  the  facts  described  are 
characterized  by  the  slope  of  the  tangent  to  the  graph.  We 
may  state  in  general : 

If  the  slope  of  the  graph  "^  is  positive,  then  the  function  in- 
creases as  the  variable  increases.     If  the  slope  of  the  graph  is 


*  The  slope  of  the  graph  is  the  same  as  the  slope  of  the  tangent. 

102 


DIFFEKENTIATION  103 

negative,  then  the  function  decreases  as  the  variable  increases. 
The  slope  of  the  graph  is  zero  at  a  maximum  or  minimum. 

39.  Differentiation.  The  discussion 
just  given  shows  plainly  that  a  method 
for  obtaining  the  slope  of  the  graph 
will  be  useful.  We  turn  our  attention 
to  this  question. 

Let  the  figure  be  the  graph  of  the 
equation  ,  ^^^^ 

(1)  y=f(x) 

It  is  required  to  find  the  slope  of  the  tangent  at  the  point 

P{x,  y). 

First  step.  Take  a  second  point  Q  on  the  curve  near  P, 
where  coordinates  ^  are  (x  +  Ax,  y  -f  A?/).  These  coordinates 
satisfy  equation  (1). 

Second  step.  Draw  PR  parallel  to  OX.  Then  BQ  =  Ay  = 
increment  of  the  function.  Also,  PR  =  Aa;  =  increment  of 
variable. 

Third  step.  Draw  the  secant  through  P  and  Q,  and  call  its 
inclination  <^.     Then,  clearly, 

tan  </)  =  tan  angle  RPQ  =  ^ , 

(2)  ,*.  tan  (fa  —  ^^  =  increment  of  function 

Ax     increment  of  variable 

Fourth  step.  Now  let  the  point  Q  move  along  the  curve 
toward  P  as  a  limiting  position.     Then 

(a)  the  secant  turns  about  P  and  approaches  the  tangent  at 
P  as  a  limiting  position ; 

(b)  the  inclination  <^  approaches  r  as  a  limit ; 

*  Ax  (read  **  delta  a;  ")  is  clearly  the  change  or  increment  in  the  value  of  x. 
Similarly,  Ay  is  the  increment  of  y. 


104  ELEMENTARY  ANALYSIS 

(c)  the  slope  of  the  secant  (=taii<^)  approaches  the  slope 
of  the  tangent  (=  tan  r)  as  a  limit. 

But  Q  will  approach  P  along  the  curve  if  we  simply  require 
that  Ai»  shall  vary  and  approach  the  limiting  value  zero.  We 
thus  obtain  from  (2), 

/o\  X  limit   Ly       ,  ^     ^^ 

(3)  tan  T  =  ^^  ^  ^  ^  =  slope  of  tangent 

The  analytical  steps  which  parallel  those  just  given  lead  to 
the  important 

General  Eule  for  Differentiation 

First  step.  In  the  function  replace  x  by  x  +  Ao?,  giving  a 
new  value  of  the  function,  y  +  Ay. 

Second  step.  Subtract  the  given  value  of  the  function  from 
the  new  value  in  order  to  find  /\y  (the  increment  of  the  function). 

Third  step.  Divide  the  remainder  Ay  (the  increment  of  the 
function)  by  Ax  (the  increment  of  the  independent  variable). 

Fourth  step.  Find  the  limit  of  this  quotient,  when  Ax  (the 
increment  of  the  independent  variable)  varies  and  approaches  the 
limit  zero. 

Let  us  apply  this  rule  to  the  equation,  p.  102, 

X 

in  which  ilf  takes  the  place  of  y. 

First  step.  M+  AM=  (x  +  Ax^  -f 


2  .      432 


x  +  Ax 
=  x^  +  2x'Ax  +  (Axf  + 


432 


x  +  Ax 


Second  step.        M  =x^  +  -^—  • 

X 


AM=2x^Ax  +  (Axf+    ^^        ^^^ 


=  (2  i»  +  Ax)  •  Ax  - 


x-{-  Ax        X 
432  Aa; 

x^  +  x '  Aa? 


DIFFERENTIATION  105 


Third  step,  =  2  a?  +  Ace 


Aa?  x^  -\-  X  '  Aa? 

Hence,  by  equation  (3),                          ♦ 
(4)  m  =  slope  of  tangent  at  {x,  y)  = ^ 

X 

Clearly,  from  this  equation,  which  we  may  write 

^2(a;^-216)^2(r^-6^) 
x^  x^ 

we  see,     if  aj  <  6,  m  <  0 ;  if  a;  >  6,  ?7i  >  0 ;  if  ic  =  6,  m  =  0, 
results  agreeing  with  the  figure. 

40.   Derivative  of  a  function.     The  general  rule  for  differ- 
entiation gives  for  any  function  y  of  x  the  value  of 


(1) 


limit   A?/ 


This  result  is  called  the  derivative  of  the  function  with  respect 
to  the  variable.  In  words  the  derivative  of  a  function  is  the 
limit  of  the  quotient  of  the  increment  of  the  function  by  the  incre- 
ment of  the  variable,  when  the  latter  increment  varies  and  ap- 
proaches the  limit  zero. 

It  is  customary  to  use  as  an  abbreviation  of  the  expression 

(1)  the  symbol  -^  (read  '^derivative  of  y  with  respect  to  a;''); 

that  is,  we  place  for  convenience 

/ox  ^_   hmit  Ay 

The  symbol  -^  is  for  the  present  to  be  regarded  as  a  whole, 
dx 

not  as  a  fraction.     In  a  later  section  we  define  dy  and  dx  sepa- 
rately and  also  -^  as  a  fraction. 
dx 


106  ELEMENTARY  ANALYSIS 

Similarly,  if  8  is  a  function  t,  then 

ds        limit  As       -i     •      i  •  n         -ii  ^  i     ^ 

—  =  .  ^     ^  —  =  derivative  of  s  with  respect  to  t. 

It  is  useful  to  write  down  symholically  the  results  of  apply- 
ing the  General  Rule  tg  the  general  equation 

(3)  y=f{x). 

We  have : 

First  step.  y  +  Ay  =f(x  +  Ax). 

Second  step.  Ay  =  f(x  +  Ax)  —  f(x) . 

Third  step.  ^  =  f(x  +  Ax)^f(x) , 

Ax  Ax 

Fourth  step.  ^  =  ?^^\  f(x-\-Ax)-f(x)  ^ 

^  dx     ^^=^  Ax 

The  derivative  of  f(x)  is  also  a  function  of  x.     We  indicate 
this  result  by  f'(x),  that  is,  we  use  the  abbreviation 

(3)  ^,^.^lun\t  f(x  +  Ax)-f(x)^ 

Summing  up,  if  y=f(x),  then 

(4)  -^=f(x)=  slope  of  tangent  at  (x,  y). 
ctx 

In  words :   the  value  of  the  derivative  of  a  function  equals  the 
slope  of  the  tangent  at  the  corresponding  point  on  the  graph. 

.    .  EXAMPLES 

1.   Differentiate  3  x^  +  5. 

Solution.     Applying  the  successive  steps  in  the  General  Bule 
we  get,  after  placing 

y  =  Sx'  +  5, 

First  step.        y  +  Ay  =  3(x+  Ax)^  +  5 

=  Sx^  +  6x'  Ax  +  3(Axy  +  5. 


DIFFERENTIATION  107 

Second  step,     y -\-  ily  =  .^  x^  -{- Q>  x  -  ^x -\-  3(Aic)^+  5 

y  =3^ +_5 

A^/  =  6  aj  •  Aic  +  3(Aa;)^  i 

Third  step,              -^  =  6  cc  +  3  •  Aa;. 
Aic 

Fourth  step,  -^  =  6x.     Ans. 

dx 

We  may  also  write  this 

^(Sx'-^5)^6x,  \    ^.     1 

dx  )  ' 

2.  Differentiate  a^  —  2  cc  -|-  7. 
Solution.     Place      y  =  x'^  —  2x-\-7. 
First  step, 

y-^Ay=(x-\-  Axf—  2(x  +  Ax-)  +  7 

=  0^34.3  a;2 .  Ao;  +  3  a; .  {Axy+(j\xy-  2  a;  -2  •  Aa;+7. 

Second  step, 

2/  +  A2/  =  o;^  +  3  a;2 .  Aa;  +  3  a; .  (Ax^  +  (Aa;)^  -  2  a;  -  2  •  Ao;  +  7 

y  =  QC^  —  2x  +7 

A2/=  3  cc^ .  Aaj  +  3  a;  .  (Aa;)^  +  (Axf  -2.Aa; 

Th  ird  step,  ^  =  3  ic^  3  a;  •  Ao; + {^xf  -  2. 

Fourth_step.  ^  =  3  ic^  _  2 .     ^ns. 

die 

Or.-^(aT^-2a;+7)  =  3a;2_2. 
do; 

3.  Differentiate  —  • 


Solution.     Place  2/  =  ^  * 


108 


ELEMENTARY  ANALYSIS 


First  step. 
Second  step, 
Tliird  step. 
Fourth  step. 


Ay  = 


(x  +  Axf 
c 


Ay 
Ax 


(x+Axy 


c  _  — c  '  Aa;(2  x+Ax) 
x^  x^{x  +  Axy 


^  =  -c 


x\x  +  Axy 
2x  2c 


Ans. 


dx  x\xy 

Or  A(±\  =  =ll. 

dx  \x^J        a^ 

4.    Eind  the  slopes  of  the  tangents  to  the  parabola  y  =  x^Sit 
the  vertex,  and  at  the  point  where  x  =  ^. 

Solution.     Differentiating  by  the  General  Rule,  we  get 

{A)  -y-  =  2x  =  slope  of  tangent  line  at  any  point  on  curve. 

ax 

To  find  slope  of  tangent  at  vertex,  substitute  a;  =  0  in  {A), 
giving 


^  =  0. 
dx 


Therefore  the  tangent  at  vertex  has  the 
slope  zero ;  that  is,  it  is  parallel  to  the  axis 
of  X  and  in  this  case  coincides  with  it. 

To  find  slope  of  tangent  at  the  point  P, 
where  x  =  \,  substituting  in  (J.),  giving 

dx 

that  is,  the  tangent  at  P  makes  an  angle  of  45°  with  the  ic-axis. 
The  derivative  may  now  be  made  use  of  in  checking  up  a 
plot. 

5.    Plot  the  locus  of 
{A)  y  =  a^-12x, 

find  the  slope  at  each   point   plotted,  and   check   up   in   the 
figure. 


DIFFEREXTIATIOjST  109 

Solution.    Differentiating  by  the  General  Rule,  we  find 
(B)  ^  =  3  0^2  -  12  =  slope  at  {x,  y). 

CtiC 


The  table  gives  the  results  of  the  calculation,  e.g.  for  x  ■ 
from    {A),   2/  =  -  11, 


1, 


and 


from    (B\    ^  = 
>    ^'    dx 


X 

y 

dy 
dx 

X 

y 

dy 
dx 

0 

0 

-12 

0 

0 

-12 

1 

-11 

-  9 

-  1 

11 

-  9 

2 

-16 

0 

-2 

16 

0 

3 

-  9 

15 

-3 

9 

16 

4 

16 

36 

-4 

-16 

36 

etc. 

etc. 

etc. 

etc. 

etc. 

etc. 

3  -  12  =  -  9  =  slope 
at  (1,  - 11). 

The  scales  used  in 
the  figure  are  those 
indicated.  To  con- 
struct the  tangent  at 
any  point,  proceed  as 
follows :  At  (0,  0),  the  slope  =  —  12.  Hence  lay  off  from  the 
origin  1  unit  to  the  left  and  12  units  up.  The  tangent  at 
(0,  0)  passes  through  this  point  (p.  18).  Similarly,  at  (3,  —9), 
slope  =  15,  hence  lay  off  1  unit  to  the  right  and  15  units  up, 
the  connecting  line  being  the  tangent. 


110  ELEMENTARY   ANALYSIS 

Note  that  different  scales  on  x  and  y  change  the  inclination, 
but  the  construction  for  the  tangent  is  clear. 

Discussion.  The  graph  has  a  maximum  point  at  (—2,  16) 
and  a  minimum  at  (2,  — 16).     There  are  no  other  horizontal 

tangents,  since  -^=z^x^  —  12  =  0  when  a;  =  ±  2,  only. 

(ajX 

PROBLEMS 

Use  the  General  Rule  in  differentiating  the  following 
examples. 

1.  2/  =  3a;l  4.,  y  =  x^.  7.  s  =  t"-2t  +  3, 

Ans.  ^  =  6x.  Ans.   ^^Sx",        A71S.    ^  =  2t-2. 

dx  dx  dt 


2.  y  =  x^-3x,  5.  r  =  ae^ 


8.  y  =  -' 

X 


Ans.  ^  =  2x-3.  Ans.  ^  =  2a^.       Ans.  ^  =  -\- 

dx  dO  dx         x^ 

3.  y=ax^+bx+c.      6.  p  =  2q\  f 

Ans.   ^  =  2ax  +  b,  Ans.    ^  =  4:0.         Ans.    ^  =  -i. 

dx  dq  dt  f 

10.  Find  the  slope  of  the  tangent  to  the  curve  y  =  2i]i^  —  6x 
+  5,  (a)  at  the  point  where  x  =  l',  (6)  at  the  point  where  x=0. 

Ans.  (a)  0 ;  (b)  -  6. 

11.  (a)  Find  the  slopes  of  the  tangents  to  the  two  curves 
y=: 3x^  —  1  and  y  =  2x^-{-3  at  their  points  of  intersection. 
(b)  At  what  angle  do  they  intersect  ? 

Ans.   (a)  ±12,  ±  8 ;  (b)  arc  tan  -/y. 
Plot  the  locus  of  each  of  the  following,  find  the  slope  at 
each  point  plotted,  and  check  up. 


12.   (a)  y  =  x'-l.  (/)  2/  =  ar^-3 


X. 


(b)  y  =  3  —  x^.  (g)  y  =  ^x^~4:X-{-l. 

(c)  y  =  2  —  4,x—x^.  (h)  y  =  ^x^  —  x^  —  3x. 

(d)  y  =  4,x-\- x^.  (i)  y  =  ^a^  —  3xF-]-5x. 

(e)  y  =  x^  —  6x,  (j)  y  =  x'^—Sx^ -{-10, 


o 


CHAPTER   VIII 


FORMULAS  FOR   DIFFERENTIATION 


41.  Theorems  on  limits.  In  the  last  step  of  the  process  of 
differentiation  the  increment  of  the  variable  is  assumed  to 
'^  vary  and  approach  the  limit  zero."  This  statement  is  made 
precise  by  the  following  definition : 

A  variable  is  said  to  approach  the  limit  zero  when  its  numerical 
value  becomes  and  remains  less  than  any  positive  number,  how- 
ever small. 

Again,  in  differentiation,  we  start  with  certain  values  of  x 
and  ?/,  and  these  values  re- 
main fixed  in  the  four  steps. 
The  actual  variables  are  clear- 
ly Aic  and  A^.  The  fact  is,  of 
course,  that  the  point  P(x,  y) 
is  fixed,  but  the  point  Q 
moves  toward  P,  and  thus  Ao; 
and  A?/  both  vary.  ]^ow  in 
the  figure,  the  position  of  Q 
depends  only  upon  the  choice  of  the  point  N,  and  the  position 
of  N  depends  upon  the  value  of  Aa*  only.  Hence  the  slope  of 
the  secant  is  a  function  of  Aic  only ;  that  is,  when  P  is  fixed, 
we  have 


(1) 


tan  a  =  ^  =  a  function  of  Lx  =  6  (Aa;). 
Aaj  ^^     ^ 


The  derivative,  that  is,  the  slope  of  the  tangent  at  P(x,  y),  is 
the  value  of 

limit    ,  /  A  ^\ 

111 


112  ELEMENTARY   ANALYSIS 

that  is,  it  is  the  limiting  value  of  a  function  of  Ace,  when  the 
variable  ^x  approaches  the  limit  zero. 

'  It  is  clear  that  the  successive  values  of  <^  (Aa^)  are  the  slopes 
of  the  successive  secants  through  P  and  the  successive  positions 
of  the  point  Q. 

How  shall  the  value  of  ^^^^  <i>  (Aa?)  be  found  ?     In  all  cases 

thus  far  the  limiting  value  has  been  found  by  substitution 
directly  in  the  function  Aa;  =  0.  In  other  words,  we  have 
assumed  the  definition : 

Tlie  limiting  value  of  a  function  of  Ax  when  Ax  varies  and 
approaches  the  limit  zero  is  the  value  of  the  function  when  Ax 
equals  zero. 

To  make  it  clear  that  difficulties  arise  in  applying  this 
definition,  consider  the  two  examples : 

limit   Va?  -{-  Ax  —  Va?  -,     limit  sin  Ax 

Direct  substitution  leads  in  each  case  to  the  meaningless  or 
indeterminant   expression    -•      In   each   of   these    examples, 

therefore,  the  definition  does  not  apply,  and  other  methods 
must  be  used  (see  Arts.  51  and  54). 

In  the  following  pages,  repeated  application  is  made  of  the 
following  theorems,  whose  proof  is  here  omitted. 

Given  a  number  of  variables  whose  limits  are  known; 
then 

I.  Hie  limit  of  an  algebraic  sum  of  any  number  of  variables 
equals  the  same  algebraic  sum  of  their  respective  limits, 

II.  The  limit  of  the  product  of  any  number  of  variables  equals 
the  product  of  their  respective  limits, 

III.  The  limit  of  a  quotient  of  two  variables  equals  the  quotient 
of  their  respective  limits  when  the  limit  of  the  denominator  is  not 
zero. 


FORMULAS   FOR   DIFFERENTIATION  113 

42.  Fundamental  formulas.  For  economy  of  time  in  differ- 
entiation, special  rules  have  been  devised,  which  are  here  set 
down,  the  proofs  being  given  in  later  sections. 

In  each  formula,  ii,  v,  w,  etc.,  are  assumed  to  be  functions  of 
the  same  variable  x,  and  a,  c,  e,  and  n  are  constants. 

I  ^  =  o. 

II  ^  =  1. 

III  JL(u  +  v-w)='^  +  §L-<^. 

doc  ddo     doc      ddo 

IT  ^(c^.)=c^- 

ddc  dx^ 

doc  doc        doo 

TI  A.(v^)=nv^-^^' 

dot)  doo 

VI  a  —  (iJC^)  =  nx^-\ 

doc 


Til 


^du_^dv 

d  fu\  _     dx^         doc 

doc  \v  )  ~  v'^ 

du 


doc  \cl       c 

VIII  ^(iog^v)=Mnedv^ 
doc  V      doc 

Villa  #^(lo^v)  =  -4^ 

doc  V  doc 

IX  ^(a^)=anoga^ 

doc  doc 

IX  a  _^(^.)^^.^. 

doc  doc 

X  ^(sini;)  =  cosi;^. 

doc  doc 


114  ELEMENTARY  ANALYSIS 

XI  :^(cosi;)  =  -sinv^. 

doc  doc 

XII  ^(tant;)  =  sec2t;^. 

doc  doc 

XIII  4-  (cot  v)  =  -  csc2 1;  ^  . 

XIV  ~  (sec  v)  =  sec  vUnv^- 
djc  doo 

XV  ^(cscv)=-cscvcoti;^. 

doD  doo 

XVI  -^  (arc  sin  v)  = ^ ^ . 

XVII  #^  (arc  tan  v)  =  -J—  ^ 

doo  1  +  t;^  c^x 

43.  Differentiation  of  a  constant.  A  function  that  is 
known  to  have  the  same  value  for  every  value  of  the  inde- 
pendent variable  is  constant,  and  we  may  denote  it  by 

y  =  c. 

As  X  takes  on  an  increment  Ax,  the  function  does  not  change 
in  value ;  that  is,  Ay  =  0,  and 

^  =  0. 

Ax 

But  ;i^^^J^V^=0. 

I  .-.^  =  0. 

dx 

Tlie  derivative  of  a  constant  is  zero. 

44.  Differentiation  of  a  variable  with  respect  to  itself. 

Let  y  =z  X. 

Following  the  General  Rule,  p.  104,  we  have 
First  step,  y  +  Ay  =  x+  Ax, 


FORMULAS   FOR  DIFFERENTIATION  115 

Second  step.  Ay  =  Ax. 

TJiird  step,  —^  =  1. 

Ax 

Fourth  step,  -^  =  1. 

dx 

II  ...^  =  1. 

Tlie  derivative  of  a  variable  with  respect  to  itself  is  unity, 

45.  Differentiation  of  a  sum. 

Let  y  =  u  +  v  —  w. 

By  the  General  Rule :  First  step.     Changing  a?  to  a?  +  Ax, 

y  ■\-  Ay  =  u-\-  Au-{-v-{-  Av  —  w  —  Aw, 
Second  step.  Ay  =  Au  ■+■  Av  —  Aw. 

Third  step.  Ay^Au_^^v_Aw^ 

Ax      Ax      Ax     Ax 

rjj      .1     .                             dy     du  ,  dv     dw 
Fourth  step.  -^  = 1 . 

dx     dx     dx     dx 
[Applying  Theorem  I,  p.  112.] 

dot)  doc     doc     doc 

Similarly  for  the  algebraic  sum   of   any  finite   number   of 
functions. 

The  derivative  of  the  algebraic  sum  of  a  finite  number  of  func- 
tions is  equal  to  the  same  algebraic  sum  of  their  derivatives. 

46,  Differentiation  of  the  product  of  a  constant  and  a 
function. 

Let  y  =  cv. 

First  step.     Changing  a:;  to  ic  +  Ax, 

y -\-  /^y  =z  c  (v  -{-  Av)  =  CV  +  cAV. 


116  ELEMENTARY   ANALYSIS 

Second  step.  Ay  =  c  -  Av. 

Third  step.  J  =  c — . 

Ax        Ax 

Fourth  step,  — ^  =  c  — . 

dx        dx 


K 


[Applying  Theorem  II,  p.  112.] 
doo  doc 


The  derivative  of  the  product  of  a  constant  and  a  function  is 
equal  to  the  product  of  the  constant  and  the  derivative  of  the 
function, 

47.   Differentiation  of  the  product  of  two  functions. 

Let  y  =  uv. 

First  step.     Changing  a?  to  a;  +  Ax, 

y-\-  Ay  =  (u-\-  Au)  {v  +  Av) 

=  uv  -{-u  '  Av  -\-v  '  Au  +  Au  •  Av. 
Second  step.  Ay  =  u  -  Av  +  v  •  Au  +  Au  .  Av. 

Third  step,  ^  =  u  —  + "": f-  Au — . 

Ax        Ax        Ax  Ax 

Fourth  step,  -^=  u [-v — . 

dx        dx        dx 

Applying  Theorem  II,  p.  112,  since  when  Ax  approaches  zero  as  a" 

limit,  Au  also  approaches  zero  as  a  liuiit,  and  limit  I  Au  —  ]  =  0. 

\       Ax/ 

Y  ..d^^^^^^dv^^^ 

doc  doc         doc 

The  derivative  of  the  product  of  two  functions  is  equal  to  the 
first  function  times  the  derivative  of  the  second,  plus  the  second 
function  times  the  derivative  of  the  first. 

To  extend  V  to  the  product  of  three  functions,  proceed  thus : 


FORMULAS   FOR  DIFFERENTIATION  117 

—  (uvw)  =  —  (uv  •  w). 
where  we  now  regard  uv  as  one  function. 

(V)  ,'.  —  (uv  -w)—  uv \-w  —  (uv^ 

^  ^  dx^  ^  dx         dx^     ^ 

dw  ,        dv  ,        du 

=  uv \-  wu \-wv . 

dx  dx  dx 

The  general  rule  to  be  read  out  of  thjs  result  is : 

The  derivative  of  the  product  of  any  number  of  functions  equals 
the  sum  of  all  the  products  that  can  be  formed  by  multiplying  the 
derivative  of  one  function  by  all  the  remaining  functions. 

48.  Differentiation  of  a  power  of  a  function.  In  equation 
(1)  of  Art.  47,  if  we  assume  ^i  and  iv  to  be  identical  with  v,  we 
obtain  at  once 

dx  dx 

which  proves  formula  VI  for  n  =  3.  In  a  similar  manner  we 
may  prove  VI  for  any  positive  integer.  For  the  present  we 
assume  VI  to  hold  if  n  is  any  constant,  proof  being  reserved 
for  a  later  section. 

In  VI,  putting  v  =  x,  we  obtain  VI  a,  using  II. 

Power  rule.  The  derivative^  of  a  function  with  a  constant 
exponent  equals  the  product  of  the  exponent,  the  function  tvith 
exponent  less  one,  and  the  derivative  of  the  function, 

49.  Differentiation  of  a  quotient. 

Let  2/=-,  '^'^0. 

V 

First  step.     Changing  a?  to  .'i?  +  Ace, 

,    .         u  +  Au 
y  +  Ay  =  —--—. 

V  -\-  Av 


118  ELEMENTARY  ANALYSIS 

V  •  Au  —  U'Av 


Second  step. 


Third  step. 


Fourth  step. 


\tl 

u-\-  Au      u 

ay  - 

V  +  Av      V 

Au         Av 
7'              ?y 

Ay 

Ax         Ax 

Ax 

v(v  4-  Av) 

dy_ 

du        dv 

V u 

dx        dx 

v(v-{-  Av) 


dx  v^ 

[Applying  Theorems  II  and  III,  p.  112.] 


VII  .-.  -f  {^\ 


(loo        doo 


The  derivative  of  a  fraction  is  equal  to  the  denominator  times 
the  derivative  of  the  numerator,  minus  the  numerator  times  the 
derivative  of  the  denominator,  all  divided  by  the  square  of  the 
denominator. 


When  the  denominator  is  constant,  set  v  =  c  in  VII,  giving 

du 

Vila  ^IVl\^^. 

doc  \c I       c 

[Since  ^i^-^  =  0.1 
L.  dx     dx        J 

We  may  also  get  VII  a  from  IV  as  follows : 

du 
d  fu\      1  du     dx 


dx\cj      c  dx       c 

The  derivative  of  the  quotient  of  a  function  by  a  constant  is 
equal  to  the  derivative  of  the  function  divided  by  the  constant. 

PROBLEMS 

Differentiate  the  following : 
1.   y  =  x\ 


FORMULAS   FOR   DIFFERENTIATION  119 

Solution.     ^=— (aj3N      3^2      ^^^^^  (by  VI  a) 

[^  =  3.] 

2.  y  =  ax^  —  bx^. 

Solution.     ^  =  -^  (ai«*  -  b^)  =  -  (aa;^)  -  ~  (ba^)       '  (by  III) 
dx     dx^  ^      dx^      ^     dx^     ^        ^  ^        ^ 

=  4  aa^  —  2  6a;.     ^ns.  (by  VI  a) 

3.  2/  =  i«^  +  5. 

Solution.     ^  =  A  (o^t)  _^  1  (5)  (by  XII) 

=  I  a;^.     ^ns.  (by  VI  a  and  I) 

3x^       7  X    ,  o  7/-0 

Solution.     ^  =  -^  (3 xV)-  ^  (7a;-3)  +  - (Sx^)  (by  III) 

da;     dx^         '     dx^        '      dx^      '  ^  •'        ' 

=  ¥  *^  + 1  «'^  +  -T-  »'"*• 

^ns.  (by  IV  and  VI  a) 

5.    y  =  (x'-3y. 

Solution.     ^  =  5(x2  -  3)*  ^_  («2  -  3)  (by  VI) 

CtX  (IX 

[ij  =  x2  —  S  and  n  =  4.] 

=  5(a;2  -  3)^  .  2  a;  =  10  x(x^  -  Sy.     Arts. 

We  might  have  expanded  this  function  by  the  Binomial 
Theorem  and  then  applied  III,  etc.,  but  the  above  process  is  to 
be  preferred.  , 


6.   y  z=  Va^  —  a^. 

Solution.     ^  =^l(a?-  x")^  =  i (a'  _  x^^  —  (a'  -  x')    (by  VI) 
dx     dx^  ^        ^  ^     dx^  J    y  J       J 

[z?  —  a^  —  x^  and  n  =  l-] 


120  ELEMENTARY   ANALYSIS 


=  i(a2  -  a^)-i(-  2x)= — — ■  •     Ans. 


7.   2/=(3a;2  +  2)Vl+5a;l 

(by  V) 


Soluti(m.     ^  =  (3  a;^  +  2)  -  (1  +  5a;^)  I  +  (1  +  6  a;^i  -  (Sa;^  +  2) 
da;  (^a;  da; 


[m  =  3  a;2  +  2  and  «  =  (1  +  5  x^)^  ] 

=  (3a;2+2)i(l +5a;^)-^— (1  +  5  a;2)  _^  (1  ^5  a^ylga; 

(by  VI,  etc.) 
=  (3  x"  +  2)(1  +  5  .T^-2  5  a;  +  6<1  +  5  x')^ 
5a?(3a)2+2)  ,  ^      /-,    ,   ^    2     45ar5  +  16a; 

a^  +  ^  ^ns. 

8.   2^= —  ' 

Solution.     ^  = 

dx  a-  —  oj- 

(by  YII) 

^2x(p?-x')-\-x(a?  +  x'') 

1 
[Multiplying  both  numerator  and  denominator  by  {aP-  —  x^)^.] 

= •     Ans. 


(a^  -  xy 
9.   2/  =  5a;^  +  3a^'-6.  ^^  =  20a;3  +  6aj. 

10.  y=*dGX^—^dx-\-5e,  -^^  =  6cx  —  Sd. 

dx 

11.  y  =  x^+\  ^  =  (a  +  b)x^+''-\ 

ctx 


FORMULAS   FOR   DIFFERENTIATION  121 

12.  y  =  x'^ -\- nx -\- n.  -^  =  nx'^~^ -{- n. 

dx 

13.  f(x)=lxf-'fx^  +  5.  f'(x)=2x^-3x. 

14.  f(x)  =  (a  +  b)x^-{-cx-]-(l         f(x)  =  2{a  +  b)x  +  c. 

15.  —  (a  +  i>^  +  c^^)  =  b  +2cx. 
dx^  ^ 

16.  —  (5  ^"*  -  3  ?/  +  6)  =  5  m?/"*-i  -  3. 

17    v  =  !^.  ^  =  _Mo. 

18.  ^  =  t'o+/^.  5=/. 

19.  s  =  So  +  Vot  +  ift'.  ^=^Vo+fi. 

dt 

20.  /z=l  +  ^>(9  +  c(9-.  —  =  h-h2cO. 

dO 

2\.    s  =  2f  +  3t-{-b.  ^  =  4^  +  3. 

22.  s  =  a/3  -  Z>^- +  c.  —  =  3at^-2bt, 

dt 

23.  r  =  a^l  ~  =  2aO, 

dO 

24.  r  =  c^3  +  ^6>^  +  e(9.  —  =  3  c^^  _|.  2  d(9 +e. 

25.  2/  =  6a;^  +  4:a?^  +  2ici         ^  =  21  a?5  +  10  x'* +  3  aji 


26.    ?/  =  V3a.'  +  A/^  + 


1  dy  _      3 


^  ^^      2V3a^      3^x^     ^ 


_  a  +  bx  +  CX^  ^hi  —  n  —  ^L 

X  dx  x^ 


122  ELEMENTARY  ANALYSIS 

1 


■gg  x''  —  x  —  x^  +  a  dy  _2  x^  +  X  +  2 x^  —  S  a 

30.  2/ =  (2 ar^  +  a;2 _ 5)3_  ^  =  6a;(3a!  +  l)(2a;'  +  a;2-5)2. 

31.  f(x)  =  (a  +  6aj2)l  y'(^)  =  ^(^^  _|.  ^^f^i, 

32.  f(x)=(l+4.a:^)(l+2x').  f\x)=4.x(l+3x-^10a^). 

33.  /(x)  =  (a  +  i»)Va-aj.  /'(i^)  = 


2Va  —  a? 

34.  /(aJ)  =  (a+aJ)"»(6+aJ)^    /'(aj)=(a  +  ^)"*(&  +  aj)"r-^H — ^1. 

[_a  +  a;     b  +  xj 

35.  2/  =  "-  ^=_JL. 

dy  __  a'*  +  a^o?^  —  4  a?* 


37       =_1^  ^^8^V-4j 

'   *^      62-aj2'  dx        (b'-xy 


38.    2/  =  ^^^=^-  ^= — 

a  +  x  dx  (a  +  xy 

39     o.^       ^  ds^Sf+l 

(1  +  0'  ^^    (1  +  0' 

41. /(^)  =  — A^.  r(e)=     "" 


Va  -  bO^  {a  -  bOf 

42.  i^(.)=Vr?|-        ^'^-)=(i-rvr^- 


FORMULAS  FOR  DIFFERENTIATION 


123 


*«=(r^)" 


43 

44.    cl>(x) 
d 


ny) 


_     my"" 


2a^-l 


X  VI  +  x^ 


<l.\x)  = 


(l_2^)-+i 


x\l  +  xy 


45. 


46. 


dx 

d 
dx 


_(a-\-x)'^(b-\-x) 


]__  m( 


(6  H-  ^)  +  ^  (<^  4-  ^) , 


"Va  +  a;  + Va 


.Va  +  ic  — V< 


a  — a?"] 


a?  ■\-  a  -\/a^  —  i»^ 


'  Va'"^  —  i 


Hint.     Rationalize  the  denominator  first. 

dx      y 
dy  Wx 


47.   y  =  -\/2  px. 


48.  y  =  -  Va^  —  x^. 

a 

49.  y  =  ((jfi  —  x^)\ 
60.    r  =  Va</)  +  c  V</>'1 


dx 


a'y 


dx  ^  X 

dr  _  Va  +  3  c<^ 


d<^ 


2V<^ 


51.    z^  = 


cc? 


dv       d  c 


52.  p  =  i2^iiD!. 


50.  Differentiation  of  a  function  of  a  function.  It  some- 
times happens  that  y,  instead  of  being  defined  directly  as  a 
function  of  x^  is  given  as  a  function  of  another  variable  v 
which  is  defined  as  a  function  of  x.  In  that  case  y  is  ^  func- 
tion of  X  through  v  and  is  called  a  function  of  a  function, 

2v 


For  example,  if 


y- 


and 


.l-x\ 


124  ELEMENTARY  ANALYSIS 

then  2/  is  a  function  of  a  function.  By  eliminating  v  we  may 
express  y  directly  as  a  function  of  x,  but  in  general  this  is  not 

the  best  plan  when  we  wish  to  find  -^  • 

dx 

Given  then 

y  =/('y)?  '^  =  <)^ (^);  and  .\y  =  F(x). 
To  differentiate,  chp.nging  a;  to  a?  +  Ao;,  then  v  and  y  become 
V  +  i^v  and  y  +  Ay  respectively.     The  third  step  will  give  the 

values  of 

Ay     Av         -,    Ay 
— ^,   — ,  and  —^^ 

Av     Ax  Ax 

But  by  direct  multiplication 

Ay      Av  __  Ay 

Av      Ax      Ax 

Taking  the  fourth  step,  we  obtain,  applying  Theorem  II, 
p.  112,  after  interchanging  the  members, 

(^)  ^^^  .  ^. 

doc     dv      doo 

If  y  =f(v)  ayid  V  —  <i>  (x),  the  derivative  of  y  with  respect  to  x 
equals  the  product  of  the  derivative  of  y  with  respect  to  v  and  the 
derivative  ofv  with  respect  to  x, 

51.  Differentiation  of  a  logarithm.  It  is  required  to  de- 
rive a  formula  for 

when  v  is  a  function  of  x,  and  the  logarithm  is  taken  in  any 
system  whose  base  is  a  positive  constant  a.  Let  us  proceed 
on  the  basis  of  Art.  50,  taking  at  first  v  for  the  variable. 

Let  y=log^v. 

First  step.     Changing  v  to  'y  +  ^^)  then 

y  +  Ay  =  log^  (v  +  Ay). 


FORMULAS  FOR  DIFFERENTIATION  125 

Second  step.  Ay  =  log„  (v  +  Av)  —  log„  v. 

TJdrd  step.  ^  =  log^(v  +  Av)~log^v^ 

Av  Av 

Fourth  step.  -^  =  - ; 

dv     0 

that  is,  the  limiting  value  of  the  right-hand  member  in  the 
third  step  cannot  be  found  by  direct  substitution  (Art.  41, 
p.  112).     We  therefore  examine  the  result  of  the  second  step, 

(2)  A2/  =  log„(^'  +  Ay)-log„v, 

and  endeavor  to  transform  the  right-hand  member.  By  15, 
p.  1,  we  may  write  (2)  in  the  form, 

(3)  A.y  =  log„(^^)  =  log.(l  +  ^). 
Dividing  by  Av,  then 

(4)  f  =  -llog/l+^' 

Av      Av         \         V 

We  observe  that  the  fraction  —  occurs  in  the  logarithm. 

V 

We  may  introduce  this  fraction  also  before  the  logarithm  by 
multiplying  and  dividing  by  v.     This  gives 

(5)  ^^l.JLlog/l  +  A!!- 

Av      V      Av         \ 

The  factor  —  in  front  of  the  logarithm  may  be  written  as  an 

Av 

exponent  on  the  parenthesis  (16,  p.  1),  and  hence 


(6)  ^  =  -l«g» 

Av      V 


1  + 


ft] 


Consider  now  the  expression  within  the   square  brackets* 
If  we  set  2;  =  — ,  this  will  have  the  form 

V 

(7)  (l+z){ 


126  ELEMENTARY   ANALYSIS 

The  fourth  step  requires  the  value  of 

limit  2 

(8)  z  =  0(l  +  z)% 

since  obviously,  when  Av  =  0,  also  z  =  —  =  0,  the  value  of  v 

V 

remaining  constant  in  differentiation  (Art.  41).  Direct  sub- 
stitution of  z  =  0  in  (7)  gives,  however,  the  meaningless  ex- 
pression 1*.  It  then  becomes  necessary  to  assume  values  of 
z  as  close  to  zero  as  we  choose  and  calculate  the  value  of  (7). 
The  limiting  value  thus  obtained,  for  example,  by  setting 

^  =  ^j  TTT?  T"o7?  TO" cm?  ^^^-y 
is  the  number  e  (Art.  33,  p.  79),  the  natural  base ;  that  is,^  as 
a  definition^  we  set 

limit  2 

(9)  e  =  z  =  0(l  +  zy. 

The  calculation  of  e  to  two  decimal  places  is  easy.  For  ex- 
ample, using  a  seven-place  table  of  common  logarithms,  we 
find  the  values 

^  —  ^  TTF  TTTO"  T170^7 

1 
(l-{.zy  =  2        2.69        2.70        2.71 

Eeturning  now  to  (6)  and  letting  Av  approach  the  limit  zero, 
we  obtain  the  required  result 

(10)  J  =  -log.e. 

dv      V 

Since  'y  is  a  function  of  x  and  it  is  required  to  differentiate 
log«  V  with  respect  to  a?,  we  must  use  formula  {A),  Art.  50,  for 
differentiating  a  function  of  a  function ;  namely, 

dy  _dy  ^  dv 
dx     dv      dx 

*  The  number  7r(=  3.1416),  with  which  the  student  is  familiar  in  geometry, 
is  defined  also  as  a  limit ;  namely,  in  a  fixed  circle,  tt  is  the  limiting  value  of 
the  ratio  of  the  perimeter  of  an  inscribed  regular  polygon  to  the  diameter, 
when  the  number  of  sides  increases  indefinitely. 


FORMULAS   FOR   DIFFERENTIATION  127 

Substituting  the  value  of  -^  from  (10) ,  we  get,  since  y  =  log«  v^ 

dv 

doc  V      ddo 

When  a  =  e  this  becomes 

since  log^  e  =  1  (19,  p.  1). 

In  VIII  a,  the  natural  base  is  omitted  in  writing  down  log  v ; 
that  is,  when  no  base  is  indicated  it  is  assumed  that  natural 
logarithms  are  used. 

Putting  a  =  10,  VIII  becomes 

(11)  —  logio  V  =  ^^^^0^  dv^Md^^ 

dx  V      dx      V  dx' 

where  M  (Art.  33,  (6))  is  the  modulus  of  the  common  system. 
Formula  VIII  a  justifies  the  introduction  of  the  number  e,  for 
in  the  Calculus  the  use  of  natural  logarithms  renders  unneces- 
sary writing  down  M,  and  this  results  in  a  great  saving  of 
labor. 

The  derivative  of  the  logarithm  of  a  function  equals  the  product 
of  the  moduhis  of  the  system,  the  reciprocal  of  the  function,  and 
the  derivative  of  the  function, 

52.  Differentiation  of  an  exponential  function.  Let  us  find 
the  slope  at  any  point  (x,  y)  on  the  exponential  curve  (Art.  33) 

(1)  y  =  e\ 
Using  natural  logarithms,  then 

(2)  x  =  \ogy. 
Differentiating  with  respect  to  y, 

(3)  ^  =  1,  by  Villa, 
^  dy     y'    ^ 

We  wish  the  value  of  —'  •    Now  if  we  had 
dx 


dy      dx  _  ^ 


128  ELEMENTARY  ANALYSIS 

differentiated  (1)  by  the  General  Bide  the  third  step  would 
have  given  the  value  of  the  ratio  -^  •     Similarly,  from  (2)  we 

should  have  found  the  value  of  —  •     But  by  direct  multipli- 
cation, ^ 

(4)  ^  .  ^  =  1 

^  ^  .  Ax     Ay        ' 

Now  take  the  fourth  step.     Then  we  have  at  once 

dy     dx 
dx     dy 
or  the  useful  formula 

^   '  dy  doo 

When  we  differentiate  (1),  we  wish  the  value  of  -^-     From 

dx 

(5),  and  (3), 

(6)  ^  =  l^'h  =  y, 
,  dx  dy 

Referring  to  (1),  we  have  the  formula 

(7)  lie)  =  e. 

In  other  words,  the  exponential  function  e""  possesses  the 
remarkable  property  of  being  its  own  derivative.  Changing 
cc  to  iJ  in  (7)  gives 

(8)  —  6^  =  e^ 
^  ^  dv  . 

From  this  result  and  (A),  Art.  50,  we  derive 

doc  doc 

The  derivative  of  the  natural  base  with  a  f auction  as  exponent 
equals  the  whole  expression  times  the  derivative  of  the  exponent. 

To  derive  IX,  make  use  of  the  equation 


FORMULAS  FOR  DIFFERENTIATION  129 

(9)  ei^""  =  a, 

which  follows  at  once  from  the  definition  of  a  logarithm  (Art. 
33,  (1)).     Hence 

(10)  a''  =  e^i^"«. 

...Aa^  =  Ae-iog«  =  et'iogaA(^loga)=e^^°s«loga  — . 
dx         dx  dx  dx 

Hence,  substituting  from  (10),  we  obtain 

IX  4^a^  =  anoga^^' 

doo  doc 

53.  Proof  of  the  power  rule.  We  may  now  complete  the 
proof  of  VI  postponed  from  Art.  48.  Since,  as  in  (9)  of  the 
preceding  section, 

then 

(1)  t)'*  =  e"i«g«. 

. ..  A  ^n  =  A  e«  log.  ^  gn  log  V  A  Ui  log  V)  =  C^^^^^^  -  —  • 

dx  dx  dx  vdx 

Substituting  from  (1),  this  becomes 

^     ,n  _  ^^^''  ^^  _         n-l  ^^ 

dx  V  dx  dx^ 

which  establishes  the  formula. 

EXAMPLES 


1.    Differentiate  y  =  log  Vl  —  x\ 

i.(l_a;2)i 

Solution.  ^  =  ^ '-—  (by  VIII  a) 

dx       (i_^)i 


x^-l 


(1  -  xy 
Ans. 


130  ELEMENTARY   ANALYSIS 

2.    Differentiate  y  —  a^\ 

Solution. 

ax 

=  6  oj  log  a  •  a^'.     Ans, 


^  =  loga.a^^  — (3a^). 
dx  dx 


(by  IX) 


3.    Differentiate 

y 

=  be''+^\ 

Solution. 

dx        dx               , 

(by  IV) 

dx 

(by  IX  a) 

y 

=  2  hxe^'+^\     Ans. 

4.    Differentiate 

=^oWJ!::- 

Solution.     Simplifying  by  means  of  17  and  15,  p.  1, 

2/  =  i[log(l  +  a;'0-log(l~a^^)]. 


d^^l 
dx     2 


■|(i+.=)  |(i-^)- 


1+x^ 


l-\-x^ 


+ 


l-a;2 

2a; 


(by  VIII  a,  etc.) 


aj2     1- 


Ans, 


PROBLEMS 

Differentiate  the  following: 

1.  2/  =  log(i»  +  a). 

2.  2/  =  log(aic+ &). 
1+x 


3.  2/  =  log 

4.  2/  =  log 


1-x 

l-M" 
1-a^* 


dy  _ 

1 

c^a; 

x+  a 

dy_ 

a 

dx 

ax-{-b 

dy  _ 

2 

dx 

1-a^ 

dy  _ 

4a; 

c?a;     1  —  a;* 


FORMULAS  FOR  DIFFERENTIATION  131 

fill 

5.  2/  =  6"^ 

6.  2/  =  e^*+^ 

7.  y  =  \og{x^  +  x), 

8.  y  =  \og{Qi^  —  2x-\-5). 

9.  2/  =  log,(2a;  +  i«3). 

10.  y  —  x\ogx.: 

11.  /(a?)  =  log  aj^. 

12.  f(x)z=\og^x. 

JO  • 

^m^.     log^x  =  (logx)3.     Use   fii-st   VI,  v=logx,   ^=3;    and  then 
VIII. 


dx 

=  ae"". 

dx 

,4g4x+5_ 

dy  _ 

2a;  +  l 

dx 

aj^  +  a; 

dy_ 

3*2-2 

dx 

ar'-2a;  +  5 

dy  _ 
dx 

,           2+3a^ 
=  ^°^»'-2.  +  x^ 

dy  _ 
dx 

=  loga5+  1. 

f\^)= 

8 

a; 

/'(«')= 

_31og2x 

13.  /(a;)=log^±^. 


a — X 


•^«=„-i^-- 


14.  f(x)  =  log  (ic  +  Vl  +  or).         f{x)  =     ^  ^        . 

Vl+aJ^ 


dy  _ 


log  a  •  a^^'e* 


cZaj 

^  =  2a;log6.^>^^ 
dx 

^  =  2log7  '(x  +  l)7^'+^\ 
dx 

^  =  -2x\ogG'e''-^\ 

dx 


132  ELEMENTARY  ANALYSIS 


19.    r  =  a^, 


20.  r  =  a^°s^. 

21.  s  =  &''^'\ 

22.  u  =  ae'^~\ 

23.  prrze*i«g«. 


24.  —  [e^(l-.T2)-]^gx(i_2a;-ic2)^ 

25.  A^^^-l^  2e^ 


=  a^  log  a. 

dr  _ 

a^^^^loga 

d(9 

0 

=  2  ^e^'+'l 

du  _ 

_  ae""^ 

dv 

2Vv 

dp  _ 
dq 

=  e^'^^^(l  +  logq). 

26.    —  (^V^)  =  :»e"^(aa;  +  2). 


tf  - 

-'^^i+e 

,x 

28. 

y^ 

-'/-' 

29. 

y  = 

e"  --  e-' 

e  4-  e-^ 

30. 

y  = 

=  a?"a^ 

^. 


f/a:     1  +  e* 

dx     2^  ^ 

dij  _        4 

-^  =  a*a?"-i  (n  +  a;  log  a). 
dx 


64.  Differentiation  of  sin  v.    Let  us  work  out 

d    . 

—  sm  ^', 
dv 

taking,  as  indicated,  v  for  the  variable. 

Let  y  =  sin  v. 


FORMULAS  FOR  DIFFERENTIATIOIST  133 

First  step.     Changing  'y  to  'y  +  Av,  then 
y-^Ay  =  sm(v  +  Av). 
Second  step.  Ay  =  sin('y  +  Av)  —  sin  v. 

If  we  proceed  without  transforming  the  right-hand  member, 
we  shall  encounter  the  same  difficulty  as  at  the  beginning  of 
Art.  51. 

Applying  formula  42,  p.  4,  by  assuming 

A  =  v  -\-  Av,  B  =  V, 

and         /.  K^  +  J5)  =  ^  +  i  Av,         ^{A  -B)  =  ^  Av, 

we  obtain  . 

Ay  =  2  cos  (v-{-  ^  Av)  sin  -J-  Av, 

Third  step.        -^  =  cos  (v  +  i-  Av)  •  '^ — , 

^         Av  V    ^^      /        ^Av 

in  which  we  have  written  Av  beneath  the  second  factor  and 
multiplied  numerator  and  denominator  by  -|-.  If  we  now  let 
A^'  approach  the  limit  zero,  and  apply  III,  p.  112,  we  see  that 
the  first  factor  approaches  cos  'y  as  a  limit,  but  that 


limit   sin  ^  Av 

Av  =  0     1  ^y 


(1) 

cannot  be  found  by  direct  substitution  (p.  112).  Let  us  there- 
fore calculate  the  value  of 

sin  a; 

y  =  — , 

into  which  (1)  transfoi^ms  if  i»  =  i  Av,  for  small  values  of  x. 
Referring  to  the  three-place  table  of  Art.  4,  and  choosing 
angles  less  than  10°,  we  see  that  sin  x  and  x  (radians)  are 
equal  to  three  decimal  places.  That  is,  for  small  values  of  x, 
y  equals  unity  very  nearly.     We  therefore  infer  that 

/o\  limit  sino^      ^ 


lU 


ELEMENTARY  ANALYSIS 


The  proof  is  the  following. 

In  the   figure,   let   x  =  2iTG  AM  =  dire  AM  \   the   radius  0 J. 
being  taken  equal  to  unity.     Then 

\^  sinx  =  FM=FM', 

tan  X  =  TM=  TM'. 


Now     M'M<  arc  M'M<  M'T+  TM. 

.  •.  2  sin  x<2x  <.2  tan  x ; 
whence,  dividing  through  by  2  sin  x, 
^  'X'         tan  X 


sm  X      sm  x\      cos  a? 
Therefore,  taking  reciprocals 

,  sm  X 


cos  0?  <  - 


<1. 


X 


Now  let  cc  approach  zero  as  a  limit ;  then,  since  cos  0  =  1, 


and  the  value  of 


smi« 


lies  between  1  and  cos  a?,  we  must  have 


limit  sm  x 


x  =  0 


=  1. 


X 


The  result  just  established  enables  us  to  complete  the  differ- 
entiation. 


Fourth  step. 


dy 

-^  =  cos  V. 
dv 


If  'y  is  a  function  of  x,  then  by  (A),  Art.  50, 

dy  dv 

dx  dx 


and  we  thus  obtain  formula 

doc  doc 

The  statement  of  the  corresponding  rules  will  now  be  left  to 
the  student. 


FORMULAS   FOR   DIFFERENTIATION  135 

55.   Differentiation  of  cos  i;. 

Let  y  =  cos  v. 

By  31,  p.  3,  this  may  be  written 

Differentiating  by  formula  X, 


dy  fir 

-^  =  oos    - 
dx  \2 


:COS 


jdx\2       J 

TT  \f       dv\ 


dv 

=  —  sm  V  — 
dx 


fsince  cos  (  -  —  ?;  j  =  sin  r?,  by  31,  p.  3- 


XI        .-.  :^(cosi;)--smt;^ 
dx  dx 

56.    Differentiation  of  tan  v. 

Let  y  =  tan  v. 

By  27,  p.  3,  this  may  be  written 
sin  v 

y  = 

COS'V 

Differentiating  by  formula  YII, 


cos  V  -r-  (sin  v)  —  sin  v  -zr-  (cos  v) 
dy  dx^        ^  dx^         ^ 

dx 


cos'^v 

cos^-y 

dv        .  „     dv 
-—  +  sm^  v  ^r- 
dx                dx 

COS^  V 

dv; 

dx 
cos^'y 

o    dv 

=  sec-'y^^- 
dx 

136  ELEMENTARY  ANALYSIS 


XII  /.  ^  (tan  V)  =  sec2  v  ^  • 

doc  doc 


57.    Proofs  of  XIII-XV.     By  setting  (using  26  and  27,  p.  3) 

,          cosv                     1  1 

coti?=-^ — ,  sec'y  = ,  csci;: 


sm  V  cos  V  sm  v 

and   applying  VII,  X,  and   XI,  we   easily   prove  the   three 
formulas  in  question.     Details  are  left  to  the  student. 

PROBLEMS 

1.  y  =  tan  Vl  —  x. 

^  =  sec^  Vr=^  A(i-x)i  (by  XII) 

dx  dx 


[i?  =  Vl  —  X.] 


:  secVl  -  X  .  i(l  -  ajp(- 1) 


sec^  Vl  —  i 


2Vl-aj 

2.  2/  =  cos^a;. 
This  may  also  be  written 

y  =  (cos  xy. 

|  =  3(cosx)^-|(cosa,)  (by  VI) 

[v  =  COS  a:  and  w  =  3.] 
=  3  cos^  cc  (  —  sin  x)  (by  XI) 

=  —  3  sin  X  cos^  a.\ 

3.  2/ =  sin  naj  sin""  cc. 

-^=^irinx~(8mxy+sm-x  —  (smnx}  (by  V) 

[z^  =  sin  fix  and  v  =  sin^x.] 

=  sin  nx  •  n  (sin  a;)"-^—  (sin  ic)  +  sin"  x  cos  7i.'r— (ncc) 

CIX  (X,X 

(by  VI  and  X) 


FORMULAS   FOR   DIFFERENTIATION  137 

=  n  sin  nx  •  sin*""^  x  cos  x-\-n  sin""  x  cos  nx 
=  n  siii*""^  X  (sin  ?iaj  cos  ic  +  cos  nx  sin  a?) 
=  n  sin**"^  a;  sin  (n  +  1)  cc. 
4:.   y  =  sec  aa;.  ^-  =  ^  sec  ax  tan  ao?. 

dv 

5.  ?/  =  tan(aa^+&).  —  =  asec^(aa;  +  6). 

Ctaj 

6.  '?/  =  sin^a?.  -^  =  sin  2  a;. 
•^  da; 

7.  y  =  cos^  a:^.  ^  z=  —  6x  cos^  a;^  sin  a?^. 
^                                                      dx 

8.  /(?/)=  sin  2  2/ cos  2/.  f\y)='\LMOs2ycosy. 

9.  i^(a;)  =  cot^  5  a;.  F'(x)  =  10  cot  5  x  cosec-  5  a?. 

10.  F{0)  =  t2in0-0,  i^'((9)  =  tan2|9. 

11.  /(<^)  =  <^  sin  (^  +  cos  <^.  /'(<^)  =  <^  cos  <jf). 

12.  /(^)  =  sin'^^cos^.  /(^)=sin2^(3cos2^  — sin^^). 

13.  r  =  acos2^.  -—  =  —  2  a  sin  2^. 


dO 


14.  8  =  a  tan  2  a 

15.  r=aVcos2^. 

16.  ?'=a(l  — cos^). 

n 

17.  ?'  =  asin^''^- 

18.  —  (log  cos  x)  =  —  tan  a;. 

19.  -|(logtana:)  =  ^. 


ds 

de~ 

;  2  a  sec- 

2d. 

dr 

a  sin 

2d 

dO 

Vcos2(9 

dr 
dO" 

a  sin  ^. 

dr 

de~ 

:asin^^( 
o 

6 

cos  3 

138  ELEMENTARY  ANALYSIS 

20.  -7-  (log  sin^  x)  =  2  cot  x. 

ClX 

tan  x~l  dy       . 

21.  y  = -^  =  sm  ic  +  cos  a?. 

sec  X  ax 


y  =  logyjj^ 


,_  +  sinaj 
22.    -      '"-^  '    ^ 

"'  ■  smx 


dy 

1 

dx 

cos  a; 

dy_ 

1 

iTT        x\ 

23.  2/  =  log  tan  (^^+2)  ^^-^^^^ 

24.  /(a;)  =  sin(a?+a)cos(a;— a),  /'(oij)  =  cos  2  a?. 

o,r     ^/  \       '    n        \  ^r/  \      cos  (log  a;) 

25.  /(cc)  =  sm  (log  X).  f\x)  = ^    ^    '^  • 

X 

26.  /(a;)  =  tan  (log  a;).  /'(a;)=?55!(l2g^. 


27.    s  =  cos-- 


.  1 

28.    r=sm-2. 


29.  p  =  sin  (cos  g). 

30.  y^e^'''^ 


ds 

a  sin  - 
t 

dt 

f 

dr 

2cos| 

dd~ 

^ 

dp 

dq  ~ 

—  sing  cos  (cos  Q'). 

dy 
dx 

e"^=«cosa'. 

^2/ 


31.  y  =  a*^^  »*=".  ^  =  Tia*^^  "^  sec^  ?ia;  log  a, 

dx 

32.  2/  =  e'^'"'"' sin  07.  -r^  =  e^°'''(cos  a;  — sin^ic). 

ctx 

di/ 

33.  y  =  e^  log  sin  a;.  •  -^  =  e^  (cot  a;  +  log  sin  a^). 


FORMULAS   FOR  DIFFERENTIATION 


139 


58.  Inverse  circular  functions.  The  definition  of  arc  sin  x 
or  sin'^oj  (read  ^^the  arc  sine  of  a?"  or  "the  inverse  sine  of  a;") 
is  the  circular  measure  of  the  angle  wJiose  sine  equals  x.  Thus, 
if  we  refer  to  the  table  of  Art.  4,  the 

arc  sin  i  =  .524^  arc  sin  0  =  0,  arc  sin  1  =  1.571,  etc. 

Clearly,  however,  since 

sin  0°  =  0,  sin  180°  =  0,  sin  360°  =  0,  etc., 

the  value  of  arc  sin  0  may  be  0,  7r(=3.14),  2  7r(=  6.28)  etc. ; 
that  is,  arc  sin  0  has  an  infinite  number  of  values.  To  avoid 
this  ambiguity,  we  make  the  convention : 

Choose  for  the  value  of  arc  sin  x  the  smallest  value  numerically. 
Thus  arc  sin  -=:  -I  not  -— -  j,  arc  sin  —  1  =  —  "^ [  not  -^ \  etc. 

The  graph  of  the  equation 

(1)  y  =  arc  sin  x, 

under  the  convention  made,  is  the  heavy 
line  of  the  figure.     For  (1)  is  the  same  as 

(2)  X  =  sin  y. 

Here  we  draw  a  sine  curve  (Art.  34) 
alo7ig  the  y-axis.  The  values  of  x  are 
comprised  between  —  1  and  + 1  inclu- 
sive, and  the  smallest  corresponding 
values    of    y    numerically,     run     from 

to  +  -  inclusive. 

2  2 

Similarly,  the  definition  of  arc  tan  x 
or  tan~^a^  is  the  circular  measure  of  the 
angle  whose  tangent  equals  x.     Referring 


-1 


X 


to  Art.  4,  arc  tan  1 ; 


arc  tan  0 : 


:  0,  arc  tan  o)  =  -  ,  etc. 

2' 


To  avoid  ambiguity,  we  ma,ke  the  same  convention  as  be- 
fore, that  is,  choose  the  smallest  value  numerically. 


140 


ELEMENTARY   ANALYSIS 


The  graph  of  the  equation 
(3)  y  —  arc  tan  a;  or  a*  =  tan  y 

is  then  the  heavy  line  of  the  figure.     We  draw  a  tangent  curve 


TT 


along  the  ^/-axis,  with  horizontal  asymptotes  y=  —  -  and  y  = 


TT 
TT 

o 

'2-                                

_^ 

X 

TT 

• 

^  •     The  values  of  x  run  from  —  (X)  to  +  oo,  and  the  smallest 
2 

corresponding  values  of  y  from  —  -  to  +  ;t- 


59.   Differentiation  of  sin-^v  and  tan-^t^.    Let 

(1)  y  —  arc  sin  v. 
Then  also 

(2)  ^'  =  sin  y. 
Differentiating  with  respect  to  y,  we  have 

dv 


dy 


:  COS  y. 


(byX) 


FORMULAS  FOR  DIFFERENTIATION  141 

.-.  f^  =  -^.  ((5),  Art.  52) 

dv     cosy 


(3) 


But  (28,  p.  3)  cos  2/  =  Vl  -  siu^y  =  Vl  -  vK        (by  (2)) 

,   dy  _       1 


dv      Vl 


Eegarding  'y  as  a  function  of  Xy  and  multiplying  both  sides 
of  (3)  by  — ,  we  obtain  (Art.  50,  (A)), 

(4)  dy^__l__d^^ 
dx     Vl  — 'y^^^' 

which  is  formula  XVI. 

The  sign  of  the  radical  in  (4)  is,  of  course,  either  4-  or  —. 
If,  however,  v  =  x,  (4)  gives  for 

/-\  '         dy  1  . 

(o )  y  =  arc  sm  x,  -^=  -^=z=  •  ' — 

^  dx   vr=^^ 

By  the  convention  made  in  Art.  58,  the  slope  of  the  curve 

(5)  in  the  figure  is  never  negative.     Hence  the  sign  of  the  radi- 
cal in  XVI  must  be  taken  positive. 

Let 

(6)  y  =  arc  tan  v. 
Then  also 

(7)  v==  tan  y. 
Differentiating  with  respect  to  y, 


—  =  see^2/-                            (by  XII) 
dy 

.-.  f/  =  ^.                      ((6),  Art.  52) 
dv     secy 

But  (28,  p.  3), 

sec^y  =  1  +  tan^?/  =  1  +  -y",          (by  (7)) 

"  dv     1  +  v'                                      -^-^ 

Multiplying  both  members  by  —  gives  XVII. 

ctx 


dy  _ 

2 

dx 

Vl-4a;2 

dx 

2 
4  +  a;' 

dy_ 

1 

142  ELEMENTARY  ANALYSIS 

PROBLEMS 

Differentiate  the  following : 

1.  y  =  arc  sin  2  x, 

2.  y  =  arc  tan  ^  cc. 

3.  y  =  arc  sm  -  •  

a  dx      Va^  -  x^ 

_        .      a;  %_      g 

.    x-\-l  dy  1 

5.  y  =  arc  sm  — ^  •  -^  =  —  —  • 

V2  ^•'^      Vl-2.T-a;2 

6.  f(x)=x^a^^^-^a^sm-'h  f  (x)  =  2V a' -  x". 

dO  3 

7.  6  =  sin-i(3r-l) 

8.  <^  =  tan'Yf^^ 

\l  —  ar 

9.  —  {x  arc  sin  x)  =  arc  sin  a;  + 
dx 

10.  p  =  e*^"'\ 


11.    ?^  =  tan"^      — 


dr 

V6r- 

-9r2 

d<i> 
dr 

1 

1  +  r' 

X 

VI- 

^ 

dp 
dq' 

gtan~^g 

du 

2 

2 

12.   2/  =  a^c  sin  (sin  a?). 
4  sin  ic 


13.    ?/  =  arctan 


3+5  cos  aj 


14.    v=arctan?4-log\/' 


dv 

e  -  e-^ 

dy  _ 
dx 

=  1. 

dy 

4 

dx 

5+3  cos  X 

dy  _ 

2  ax' 

dx 

x^-a"" 

FORMULAS   FOR  DIFFERENTIATION  143 


15.    2/=log^l±^Y--arctana..        ^  =  ^. 


dy  _      X  arc  sin  a; 


16.   ?/  =  Vl  —  x^  arc  sin  x—x,  J~  ~" ,, 


17.   2/  =  arc  sm  — 


c7?/  _  2  710?""^ 


ic-"  +  l  ^a;      a,^^^+l 


^^      d  —\dv 

18.  —  arc  cos  v  =  — m^ziiz  —  • 
dx  ^1  _  ^^2  c^a? 

,  ^      d  ,  —  1  c??; 

19.  — arc  cot 'y  = — 

dx  1  4-  ^!'^  dx 

20.  — 860-^?;  = ■ 

dx  'y  V'y^  —.\dx 


60.    Implicit  functions.     Required  the  slope  of  the  curve 
(1)  0^  -  2  2/'  =  9. 

If  we  solve  for  y^  we  shall  obtain  ?/  as  a  function  of  x^  namely, 


(2)  y^4 


fa^-9 


Instead  of  solving  for  ?y,  however,  we  may  in  (1)  regard  y 
implicitly  as  a  function  of  x,  and  differentiate  directly.  Thus,  by 
III,  we  have  from  (1), 

Remembering  that  y  is  &  function  of  x,  then 

fj^^f)='^^/  =  ^y%  (by  VI).     Hence  (3)  gives 

(4)  2a.-4^^  =  0,or^  =  -^. 

da;  dx      2  y 


144  ELEMENTARY  ANALYSIS 

To  see  that  this  result  is  the  same  as  would  be  obtained  by 
differentiating  (2),  using  VI  in  (2), 

,..         dy     l/aj2-9V*      dfx'-^\  X  x    .     ,^. 

^'^   rx=2[-^)  'rx{r2-rrj^_=2y^^^^'^' 

The  general  conclusion  illustrated  by  this  discussion  is  the 
following : 

When  the  equation  of  a  curve  is  given  in  unsolved  form,  either 
coordinate  may  he  considered  an  implicit  function  of  the  other. 
We  may  then  differentiate  directly,  and  solve  for  the  desired 
derivative. 

Thus,  to  find  -^  from 
dx 

a?-3xy-^2y^  =  3. 

Then  -f  {x')  -3^(xy)  +  2^ {f)  =  -f  3. 

dx  dx  dx  dx 

,'.2x-3(y^-x^\-\-4.y^  =  0,  (V  and  VI) 

y  dxj  dx 


c,  T   .                              dy _2x 
^'                         dx     6x 

-4.y' 

PROBLEMS 

Differentiate  the  following : 

1.    y^  =  4:px. 

dy  _2p  ^ 
dx      y 

2.    a^  +  y''  =  r\ 

dy  _      X 
dx         y 

3.    bV  +  ay  =  a'b\ 

dy  __      Wx  ^ 
dx         a^y 

4.   f-3y-{-2ax  =  0. 

dy          2  a 
dx     3(1- f) 

FORMULAS  FOR  DIFFERENTIATION  145 

5.   xl+y^  =  a^.  ^  =  ^Jy. 

rim*  ^  /v» 


dx 

6.   a;«  +  r  =  ai  ^  =  -Ji. 

dx  ^x 

8.   /-2a;«/  +  62  =  0.  'M=jy_. 

dx     y  —  X 

dx     y^  —  ax 

dx     x^  —  xylogx 
Hint.    Take  the  logarithm  of  both  members. 

11.  p^  =  a^cos2^.  dp^_a'sm2$^ 

de  p 

12.  p' cos e  =  a' sin 3  0.  d^^'Sa'cosSO  +  p'sm  0_ 

de  2  p  cose 

13.  cos  {uv)  =  cv.  dM^c  +  usm{^iv)^ 

dv         V  sin  (uv) 

14.  ^  =  cos(^4-<i).  ^  = sin(^  +  <^)     . 

d^         1  + sin  (61  +  .^) 


CHAPTER   IX 

SLOPE,    TANGENT,   AND   NORMAL 

61.  If  the  equation  of  a  curve  is  given  in  rectangular  coor- 
dinates, it  has  been  shown  in  Art.  40,  p.  106,  that 

(1)  ^  =  tan  a  =  slope  of  tangent  (or  curve)  at  {pc,  y)  =  ^in. 
(too 

To  find  the  slope  at  any  particular  point  we  have  merely  to 
substitute  the  coordinates  of  that  point  into  the  expression  for 
the  derivative  (as  on  p.  109).  Having  thus  found  the  numeri- 
cal value  of  m,  the  equation  of  the  tangent  is  given  by  the 
point-slope  form,  Art.  27, 

(2)  y-ij.^mix-Xi), 

the  point  of  contact  being  (x^,  t/i)- 

The  normal  is  the  perpendicular  to  the  tangent  drawn 
through  the  point  of  contact.  We  therefore  find  its  equation 
by  the  formula 


(3)  S-H,' 


m 


■  ^i)y 


where  m  =  slope  of  the  tangent. 


EXAMPLES 


Given  the  curve  y  =  —  - 
^      3 


2  (see 


figure). 

(a)  Find  a  when  x  =  l. 

(b)  Find  a  when  x  =  3. 

(c)  Find  the  points  where  the  curve  is 
parallel  to  OX. 


(d)  Find  the  points  where  a  =  45°.. 

146 


SLOPE,   TANGENT,   AND  NORMAL 


147 


(e)  Find  the  points  where  the  curve  is  parallel  to  the  line 
2x-dy  =  ^  (line  ^5). 

Solution.      Differentiating,     -^  =  a;^  —  2  a?  =  slope    at    any 
point  =  tan  a, 

(a)  tan  a  = 
(h)  tan  a  = 


dy 
dx 

dy 
dx 


=  1  -  2  =  -  1 :  therefore  a  =  135^     Ans. 


:  9  —  6  =  3 :  therefore  a  =  arc  tan  3.   Ans. 


(c)  a  =  0°,  tan  a  =  0 ;  therefore  a;^  —  2  a;  =  0.  Solving  this 
equation,  we  find  that  x  =  0  or  2,  giving  points  C  and  D  where 
curve  (or  tangent)  is  parallel  to  OX. 

(d)  a  =  45°,  tan  a  =  1 ;  therefore  x-  —  2  a;  =  1.  Solving,  we 
get  aj  =  1  ±  ■\/2,  giving  two  points  where  the  slope  of  curve  (or 
tangent)  is  unity. 

(e)  Slope  of  line  =  | ;  therefore  of  —  2  x  =  ^.  Solving,  we 
get  x  =  l±  V|,  giving  points  E  and  F  where  curve  (or  tan- 
gent) is  parallel  to  line  AB. 

Since  a  curve  at  any  point  has  the  same  direction  as  its  tan- 
gent at  that  point,  the  angle  between  two  curves  at  a  common 
point  will  he  the  angle  between  their  tangents  at  that  point. 

2.   Find  the  angle  of  intersection  of  the  circles 

(A)  x^  +  y^-4.x  =  l, 

(B)  x'  +  y2_2y  =  9. 

Solution.  Solving  simultaneously,  we 
find  the  points  of  intersection  to  be  (3,  2) 
and  (1,  -2). 

|  =  ^fi'om  (A),  (by  §  60,  p.  144) 


dy 


dx     1 


from  {B).  (by  §  60,  p.  144) 


148 


ELEMENTARY  ANALYSIS 


r2-x~ 

y  - 

~      X      ~ 

y=2 

l^-yj 

x-3 
y=2 

=  —  -  =  slope  of  tangent  to  (A)  at  (3,  2). 


:  —  3  =  slope  of  tangent  to  (B)  at  (3,  2). 


From  p.  69,  we  know  that  the  locus  of  (A)  is  a  circle  with 
center  at  (2,  0)  and  radius  =  VB,  and  of  (B)  also  a  circle  with 
center  at  (0,  1)  and  radius  =  VlO. 

The  formula  for  finding  the  angle  between  two  lines  whose 
slopes  are  mi  and  mg  is 


tan  6  —  — - 


mo 


1  +  ^^1^2 


((IV),  Art.  29) 


To  find  the  smaller  angle  of  the  figure,  set  (p.  65)  mi  =  —  i, 
m2  =  — 3. 

Substituting,  tan  0  = 


-^  +  3. 


1  +  f 


:  1 ;  therefore  0  =  45°.     Ans. 


This  is  also  the  angle  of  intersection  at  the  point  (1,  —2). 

3.  Find  the  equations  of  the  tangent  and  normal  to  the 
parabola  y^  =  4:X  +  l  at  the  point  where  x  =  2  and  the  ordinate 
is  positive. 

Solution.     Substituting  x  =  2,  then  2/^=9,  y  =  ±i  3,  and  the 

point  of  contact  is  (2,  3). 

Differentiating  (Art.  60),  we  have 

—  =  -  =  slope  at  (Xf  y). 
dx     y 

,\  slope  at  (2,  3)  =  |  =  m. 

Hence  the  required  equations  are 

PT:y-3  =  i(x-2),ov 

2x-Sy  +  5  =  0', 

PN:y-3=  -  ^(x-2),  ov  3  x  +  2  y -12  =  0. 


SLOPE,   TANGENT,   AND  NORMAL  149 

PROBLEMS 

The  corresponding  figure  should  be  drawn  in  each  of  the 
following  problems. 

1.  Find  the  slope  of  y  =  — ^^  ^^  (^?  ^)'        ^^^^-  ^  =  ^^^^  ^' 

2.  What  is  the  direction  in  which  the  point  generating  the 
graph  of  y  =  3x^  —  X  tends  to  move  at  the  instant  when  x  =  l? 

Ans.  Parallel  to  a  line  whose  slope  is  5. 

3.  Find  the  points  where  the  curve  y=.x^  —  3x^  —  9x  +  5 
is  parallel  to  the  axis  of  X,  Ans.  x  =  3  and  x  =  l. 

4.  Find  the  points  where  the  curve  y  (x  —  1)  (cc  —  2)  =  a;  —  3 
is  parallel  to  the  axis  of  X.  Ans.  x  =  3  ±  V2. 

5.  At  what  point  on  y^  =  2x^  is  the  slope  equal  to  3  ? 

A71S.  (2,4). 

6.  At  what  points  on  the  circle  x'^-\-y^=  i^  is  the  slope  of 
tangent  line  equal  to  —  |  ?  Ans.   ±  f  r,  ±  f  r. 

7.  Where  is  the  tangent  to  the  parabola  y=zx^  —  lx-\-3 
parallel  to  the  line  ?/  =  5  a;  +  2  ?  Ans.  (6,  —  3). 

8.  Find  the  points  where  the  tangent  to  the  circle  x^  +  y^ 
=  169  is  perpendicular  to  the  line  bx-^12y  =  60. 

'  Ans.  (±12,  ±5). 

9.  At  what  angles  does  the  line  3y  —  2x  —  S  —  0  cut  the 
parabola  y^  =  Sx?  Ans.  arc  tan  i,  and  arc  tan  i 

10.  Find  the  angle   of  intersection  between   the   parabola 
y^  =  6x  and  the  circle  x^  +  2/^  =  16.  Ans.  arc  tan  |  V3. 

11.  Show   that  the   hyperbola  x^  —  y^=5   and  the  ellipse 

OCT  V^ 

—  +  ^  =  1  intersect  at  right  angles, 
lo       o 

12.  At  how  many  points  can  the  curve  y  =  x^  —  2x'  +  x  —  4: 
be  parallel  to  the  axis  of  X  ?     What  are  the  points  ? 

Ans.  Two;  at  (1,  -4)  and  (i   -VV)- 

13.  Find  the  angle  at  which  the  parabolas  2/ =  3  i»^  —  1  and 
y  =  2x^-]-3  intersect.  A71S.  arc  tan  -^j. 


150  ELEMENTARY   ANALYSIS 

14.    Find  the  equations  of  the  tangent  and  normal  to  each 
of  the  following  curves  at  the  point  indicated. 


(a)  x^  +  f  =  2^, 

(3,  4). 

Ans. 

3x-i-4.y  =  25, 
4:X-3y  =  0. 

(b)  y'  =  2x  +  5, 

(2,-3). 

Ans. 

x  +  3y  +  7=0, 
3x-y-9  =  0. 

(c)  y  =  a^-5, 

(2,  3). 

Ans. 

12a?-2/-21  =  0, 
07  +  12  2/-38  =  0. 

(d)  42/2  =  0^  +  8, 

(1,  -f). 

Ans. 

(e)  x'-Af=:12, 

(-4,1). 

Ans. 

x  +  y  +  3  =  0, 

x  —  y-{-5  =  0. 

(/)  2/  =  2  sin  X, 

(0,  0). 

Ans. 

2x-y  =  0, 
x  +  2y  =  0. 

(g)  y  =  e% 

(0,  1). 

Ans. 

x-y  +  l  =  0, 
x  +  y-l  =  0. 

(h)  x'  +  f  =  r', 

(^1,  Vi)' 

Ans. 

yj,x-x^y  =  0. 

(i)  y^  =  2px, 

(^1,  yi)' 

Ans. 

yiy=p{x  +  x^, 

y-yi=--(^-^i)' 
p 

(j)  y  =  o(^,  (xy,  2/i).  Ans.  3  x^x -y-2y^  =  0, 

x  +  3  x^hj  -  x^{3  x^^  + 1)  =  0. 

\   15.   Plot  each  of  the  following  curves,  find  the  slope  at  each 
pVint  plotted  (p.  109),  and  locate  all  horizontal  tangents, 
(a)  x^  +  xy  +  S  =  0. 
\h)  y^-2xy-4:  =  0. 
^  (c)  x^y-5  =  0. 
X—  3 


id)y-. 


(f)y  = 


x  +  l 


X 

X 


1  +  x' 
(g)  y^xe-". 


Qi) 

y  = 

e-< 

(0 

y  = 

a^e-^ 

(J) 

y  = 

:  X  log  X. 

{k) 

y  = 

■■  log  a;  -=-  X. 

(J) 

y  = 

X  +  sin  X. 

(m) 

y  = 

\x  —  cos  X. 

(n) 

2/  = 

X  —  sin  2  X. 

(0) 

y  = 

■■  sin  X  +  cos  X. 

(P) 

y  = 

■  sin  X  +  sin  2  x. 

(9) 

y  = 

cos  ^  7705  —  COS  |-  TTX, 

CHAPTER   X 


MAXIMA   AND   MINIMA 

62.  The  graphs  of  the  functions  in  Chapter  VI  exhibited 
"  high  points  "  or  "  low  points ''  corresponding  respectively  to 
maximum  or  minimum  values  of  the  function. 

A  maximum  value  of  a  function  is  a  value  which  is  greater 
than  all  values  immediately  preceding  or  following. 

A  minimuin  value  of  a  function  is  a  value  which  is  less  than 
all  values  immediately  preceding  or  following. 

Clearly,  2i first  condition  for  a  maximum  or  minimum  value 
is  that  the  slope  of  the  graph  shall  be  zero.  That  is,  values 
of  the  variable  for  which  such  values  of  the  function  f{x) 
occur,  satisfy 

(1)  /(x)=0. 

All  such  values  of  the  variable  are  called  critical  values. 

The  statement  of  the  problem  often  indicates  whether  a 
critical  value  corresponds  to  a  maximum  or  a  minimum.  This 
was  seen  to  be  the  case  in  the  problems  given,  p.  96.  It  is  easy 
to  check  up  this  foreseen  result.  For,  clearly,  at  a  high  point 
(a)  the  slope  is  positive  just  before  and  negative  just  after  the 
critical  value,  while  reverse  conditions  hold  for  a  low  point  (h). 


The  significant  fact  concerning  maximum  and  minimum  val- 
ues of  a  function  is,  therefore,  that  the  derivative  changes  sign. 
These  results  are  summarized  as  follows : 

161 


152 


ELEMENTARY    ANALYSIS 


Determination  of  maximum  and  minimum  values. 

First  step.  From  the  given  statement  find  the  function  (Chap- 
ter VI)  and  decide,  if  possible,  whether  maximum  or  minimum 
values  exist. 

Second  step.  Differentiate,  set  the  derivative  equal  to  zero,  and 
solve  for  all  real  values  of  the  variable.  These  are  the  critical 
values. 

Third  step.  Consider  any  critical  value  not  excluded  by  the 
conditions  of  the  problem,  and  find  the  slope  of  the  graph  just 
before  and  just  after  the  corresponding  point.  If  the  slope  changes 
from  -\-  to  —,  the  function  is  a  maximum;  if  the  reverse  is  true^ 
the  function  is  a  minimum. 

The  third  step  is  to  be  regarded  as  a  verification  of  what  is 
foreseen  in  the  discussion  of  the  first  step. 

EXAMPLE 

Given  a  circle  of  diameter  5  in.  Examine  the  areas  of  in- 
scribed rectangles  for  maxima  and 
minima  (see  p.  92). 

Solution.  First  step.  If  x  rep- 
resents the  base  of  any  rectangle 
of  area  A,  then  (1),  p.  92, 


Inches 


dA 


(1)  A  =  x-yj2b-x\ 

Discussion  shows  that  a  maxi- 
mum exists. 

Second  step.     Differentiating, 

d 


^^  =  V25  -x'  +  x—m-x'y, 
dx  dx 


(2) 


=  V25— a!^- 


dA      25-2  a;2 


V25  -  a?' 


dx      V25  -  x^ 


MAXIMA   AND   MINIMA  153 

Setting  this  equal  to  zero,  we  obtain  25  —  2x^  =  0,  and  hence 
5 
the  critical  value  is  a;  =  — ^  =  3.55,  the  negative  value  being 

v  2 
excluded. 

Third  step.     For  a  value  just  before  the  critical  value,  use 

x  =  3.     Then  from  (2),  ^^^^ 

greater,  use  a?  =  4.     Then 


dx 
dA 


=  +  number.     For  a  value  just 

a-=3 


^^  Jx=4 


:  —  number.     Thus  the  re- 


sult foreseen  in  the  first  step  is  verified.     That  is,  the  maxi- 
mum   rectangle    has    the    area   A  =  — ^-\  /25  —  _  =  _  =  12^ 

V2>'  2       2  ^ 

sq.  in.     It  is  easy  to  see  that  the  rectangle  .is  now  a  square. 

For  when  x  =  — -,  the  altitude  =  ^25  —  x^  =  — -  (ry  w\  =  x 
V2  V2  ^^*      ^ 

Hence  this  example  proves  the  result :  — 

Of  all  rectangles  which  can  be  inscribed  in  a  given  circle, 

the  square  has  the  greatest  area. 


PROBLEMS 

In  the  following  problems  the  student  will  work  out  the 
functional  relation,  and  examine  this  for  maxima  and  minima. 

1.  Eectangles  are  inscribed  in  a  circle  of  radius  r.  Ex- 
amine the  perimeter  P  of  the  rectangles  as  a  function  of  the 
breadth  x.  Arts.  Max.  for  x  —  r  -\/2, 

2.  Right  triangles  are  constructed  on  a  line  of  given  length 
li  as  hypotenuse.  Examine  (a)  the  area  A^  and  (6)  the 
perimeter  P  as  a  function  of  the  length  x  of  one  leg. 

Ans,  (a)  Max.  for  x  =  ^  ^ V2.     (b)  Max.  f or  aj  =  i  7i  V2. 

3.  Eight  cylinders  are  inscribed  in  a  sphere  of  radius  r. 
Examine  as  functions  of  the  altitude  x  of  the  cylinder, 
(a)  volume  V  of  the  cylinder,  (b)  curved  surface  S. 

Ans.  (a)  Max.  for  x  =  ^r  V3.     (b)  Max.  for  x=zr V2. 


154  ELEMENTARY   ANALYSIS 

4.  Eight  cones  are  inscribed  in  a  sphere  of  radius  r.  Ex- 
amine as  functions  of  the  altitude  x  of  the  cone,  (a)  volume  V 
of  the  cone,  (b)  curved  surface  S, 

Ans.  (a)  Max.  if  a;  =  |  r.     (b)  Max.  if  a?  =  |  r. 

5.  Right  cylinders  are  inscribed  in  a  given  right  cone.  If 
the  height  of  the  cone  is  h,  and  the  radius  of  the  base  r,  ex- 
amine (a)  the  volume  Fof  the  cylinder,  (b)  the  curved  surface 
S,  (c)  the  entire  surface  T,  as  functions  of  the  altitude  x  of  the 
cylinder.  Ans.  (a)  Max.  ii  x  =  ^  li.     (b)  Max.  ii  x=^  1i, 

6.  Right  cones  are  circumscribed  about  a  sphere  of  radius 
r.  Examine  as  a  function  of  the  altitude  aj  of  the  cylinder,  the 
volume  Fof  the  cone.  Ans.  Min.  if  0^  =  4  r. 

7.  Right  cones  are  constructed  with  a  given  slant  height 
L.  Examine  as  functions  of  the  altitude  x  of  the  cone,  (a)  the 
volume  V  of  the  cone,  (b)  the  curved  surface  S,  (c)  the  entire 
surface  T.         Ans.  (a)  F=Max.  if  a7  =  iX  V3.     (b)  ISTeither. 

8.  A  conical  tent  is  to  be  constructed  of  given  volume  V. 
Examine  the  amount  A  of  canvas  required  as  a  function  of  the 
radius  x  of  the  base.  Ans.  Min.  if  x  =  -|-  V2  altitude. 

9.  A  cylindrical  tin  can  is  to  be  constructed  of  given 
volume  V.  Examine  the  amount  A  of  tin  required  as  a  func- 
tion of  the  radius  x  of  the  can.  Ans.  Min.  if  x  =  altitude. 

10.  An  open  box  is  to  be  made  from  a  sheet  of  pasteboard 
)     12    in.    square,    by    cutting    equal    squares    from    the    four 

corners  and  bending  up  the  sides.     Examine  the  volume  V  as 
[     a  function  of  the  side  x  of  the  square  cut  out. 

Ans.  Max.  if  a?  =  3. 

11.  The  strength  of  a  rectangular  beam  is  proportional  to 
the  product  of  the  cross  section  by  the  square  of  the  depth. 
Examine  the  strength  /S  as  a  function  of  the  depth  x  for  beams 
which  are  cut  from  a  log  12  in.  in  diameter. 

A71S.  Max.  if  X  =  6  V3. 


MAXIMA   AND   MINIMA  155 

12.  A  rectangular  stockade  is  to  be  built  to  contain  a  cer- 
tain area  A.  A  stone  wall  already  constructed  is  available  for 
one  of  the  sides.  Examine  the  length  L  of  the  wall  to  be 
built  as  a  function  of  the  length  x  of  the  side  of  the  rectangle 
parallel  to  the  wall.  A71S,  Min.  if  x  =  twice  other  side. 

13.  A  tower  is  100  ft.  high.  Examine  the  angle  a  sub- 
tended by  the  tower  at  a  point  on  the  ground  as  a  function  of 
the  distance  x  from  the  foot  of  the  tower.  Ans.  Neither. 

14.  A  tower  50  ft.  high  is  surmounted  by  a  statue  10  ft. 
high.  If  an  observer's  eyes  are  5  ft.  above  the  ground,  ex- 
amine the  angle  a  subtended  by  the  statue  as  a  function  of  the 
observer's  distance  x  from  the  tower. 

Ans.  Max.  if  a;  =  10  V30. 

15.  A  line  is  drawn  through  a  fixed  point  (a,  b).  Examine, 
as  a  function  of  the  intercept  on  XX'  {=x)  of  the  line,  the 
area  A  of  the  triangle  formed  with  the  coordinate  axes. 

Ans.  Min.  when  x  =  2  a. 

16.  A  ship  is  41  mi.  due  north  of  a  second  ship.  The  first 
sails  south  at  the  rate  of  8  mi.  an  hour,  the  second  east  at  the 
rate  of  10  mi.  an  hour.  Examine  their  distance  d  apart  as  a 
function  of  the  time  t  which  has  elapsed  since  they  were  in 
the  position  given.  Ans.  Min.  if  t=  —  2. 

17.  Examine  the  distance  e  from  the  point  (4,  0)  to  the 
points  (x,  y)  on  the  parabola  ?/^  =  4  cc.  Ans.  Min.  if  a;  =  2. 

18.  A  gutter  is  to  be  constructed  whose  cross  section  is  a 
broken  line  made  up  of  three  pieces  each  4  in.  long,  the  middle 
piece  being  horizontal,  and  the  two  sides  being  equally  inclined. 
Examine  the  area  ^  of  a  cross  section  of  the  gutter  as  a  func- 
tion of  the  width  x  of  the  gutter  across  the  top. 

Ans.  Max.  for  a?  =  8. 

19.  A  Norman  window  consists  of  a  rectangle  surmounted 
by  a  semicircle.  Given  the  perimeter  P,  examine  the  area  A  as 
a  function  of  the  width  x.       Ans.  Max.  when  x  =  total  height. 


156  ELEMENTARY   ANALYSIS 

20.  A  person  in  a  boat  9  mi.  from  the  nearest  point  of  the 
beach  wishes  to  reach  a  place  15  mi.  from  that  point  along 
the  shore.  He  can  row  at  the  rate  of  4  mi.  an  hour  and  walk 
at  the  rate  of  5  mi.  an  hour.  The  time  it  takes  him  to  reach 
his  destination  depends  on  the  place  at  which  he  lands.  Ex- 
amine the  time.        Ans.  Min.  if  he  lands  3  mi.  from  the  camp. 

21.  The  illumination  of  a  plane  surface  by  a  luminous  point 
varies  directly  as  the  cosine  of  the  angle  of  incidence,  and  in- 
versely as  the  square  of  the  'distance  from  the  surface.  Ex- 
amine the  illumination  /of  a  point  on  the  floor  10  ft.  from  the 
wall,  as  a  function  of  the  height  a;  of  a  gas  burner  on  the  wall. 

Ans,  Max.  if  aj  =  5  V^. 

22.  The  sides  of  a  quadrilateral  are  given  in  order  and 
length.     When  is  the  area  a  maximum  ? 

23.  Examine  the  functions  of  problems  22-30,  p.  99,  for 
maxima  and  minima. 

€3.  Derivatives  of  higher  orders.  Since  the  derivative  of 
a  function  of  a  variable  x  with  respect  to  x  is  also  in  general  a 
function  of  x^  we  may  differentiate  the  derivative  itself,  that 
is,  carry  out  the  operation, 

This  double  operation  is  indicated  by  the  more  compact 
notation, 

and  this  new  function  is  called  the  second  derivative.     In  the 

same  way,  ^    ^2  ^3 

f(x)  =  —  f(x) 

dx  dx"^  ^  ^      dx^^  ^  ^ 

is  the  third  derivative,  and  in  general, 

dx-^  ^  ^ 


MAXIMA  AND   MINIMA  157 

is  the  nth.  derivative  of  f{x),  that  is,  the  result  of  differentiat- 
ing/(a;)  n  times.     The  following  notation  is  also  used: 

1/(0.)  =/'(^),  £jix)=r(x), ...,  £./(«,)=/(».(«,). 

The  operation  of  finding  the  successive  derivatives  is  called 
successive  differentiation. 

For  example, 

given  f(x)  =  3  o^^  —  4  o?^  +  6  a;  —  1, 

then  /(a;)=12ar^-8a;  +  6; 

/'(a^)  =  36a;'^-8,  etc. 

If  the  independent  variable  is  y,  then  the  second  differentia- 
tion gives  the  value  of 

d  fdy\  _  d'^y 
dx\dxj  ~~  dx^^ 

the   abbreviation   indicated   being   that  commonly  employed. 

The  symbol  — ^  is  read,  "  the  second  derivative  of  y  with  respect 

(JjX" 


X."     isimiiariy, 

d  (d'y\_d'y 
dx\dxy~da^' 

the  third  derivative. 

etc. 

Thus,  given 

y  =  x^  —  4:X^. 

Then 

^  =  Sx^-8x; 
dx 

2  =  6^     8; 
d3^ 

^  =  6; 

d3i> 

fy  =  0;  etc. 
dx* 

J    y^j  - 

(1- 

-xf 

r\y)  = 

=  16. 

d*y_ 

6 

dx* 

X 

dhj  _ 

_  11  (n 

+  l)c 

158  ELEMENTARY  ANALYSIS 

PROBLEMS 

Differentiate  the  following. 

r^  14 

3.  /(^)=/. 

4.  y=^oi?  log  X, 


G 

5.  V  "= — • 

6.  ?/=(aj-3)e2^  +  4a;e^  +  a;.      ^  =  4e^[(a;- 2)e*  +  a;  +  2]. 

cix 

8.  /(«)  =  a*- +  &«  +  c.  /"'(a;)  =  0. 

9.  f(x)  =  log  (a;  +  1).  /'  (x)  =  -  -^^ . 

10.  /(a;)  =  log  if  +  e-).  /"  (a;)  =  -  ^/f~g  " 

11.  r=sina^.  — -  =  a^  sin  a^  =  aV. 

12.  r  =  tan^.  -^  =  6  sec^  <^  -  4  sec^  <^. 

d^ 

dh 

d<t>^ 

14.  /(^)  =  e-'  cos  ^.  /^(f)  =  —  4  e-*cos  ^  =  —  4/(f). 

15.  /(^)=Vse^T^.  r(^)=3[/(^)]^-/(^). 


13.   r  =  logsin<^.  — ^  =  2  cot  (^  cosec^  ^. 


64.   Geometrical  significance  of  the  second  derivative.    It 

has  been  observed  that  the  value  of  the  first  derivative  of  a 


MAXIMA  AND   MINIMA 


159 


function  determines  the  slope  of  the  graph.      Moreover,  we 
have  seen,  in  Art.  38,  that 

(i)  if  the  derivative  is  positive^  the  function  increases  as  the 
variable  increases ; 

(ii)  if  the  derivative  is  negative ,  the  function  decreases  as  the 
variable  increases. 

If  we  consider  the  appearance  of  a 
curve  in  the  neighborhood  of  one  of 
its  points  P,  then  clearly  two  cases 
are  distinguishable : 

(a)  The  curve  is  above  the  tangent  at  P,  or  is  concave  upward; 
(h)  The  curve  is  below  the  tangent  at  P,  or  is  concave  down- 
ward. 

How  are  these  cases  distinguishable  analytically  ? 

Consider  (a).  As  we  approach  P 
from  the  left,  that  is,  with  increasing 
values  of  x,  we  observe  from  the  fig- 
ure that  the  inclination  a  of  the  tan- 
gent increases  (a' >  a).  That  is,  in 
(a),  the  inclination  a  increases  as  x 

increases.  Hence  by  (i),  if  —  is  pos- 
itive at  P,  the  curve  is  certainly  con- 
cave upward. 

Consider   (b).      Here,    a   decreases 
as   X  increases    («'<a).      Hence   if 
da 
dx 


%> 


X' 


<  0,  the  curve  is  concave  downward. 
Summing  up,  if 


(A)  —  >  0,  curve  is  concave  upward;  —  <  0,  concave  downward, 
dx  dx 

N"ow,  \if(x)  is  the  derivative  of  the  function /(o:^),  then 

(2)  f\^)  =  tan  a,  or  a  =  arc  tan  /'  (x). 


16Q 


ELEMENTARY   ANALYSIS 


Differentiating  with  respect  to  x,  using  XVII, 

,ox  d(._  dx^^^''^  ^  r(x) 

^^  dx     l-{-f\x)     l+f\xy 

where /"  (a?)  =  second  derivative  oif{x).     Hence  (3)  becomes 
da_     f\x) 


(4) 


dx  i-vr\x) 


da 


Since  the  denominator  l+/'^(a?)  is  always  positive,  —  has 

the  same  sign  as/"(a;),  and  conversely.     Hence  the  result : 
second  derivative  >  0,  curve  concave  upward ; 
second  derivative  <  0,  curve  concave  downward. 

As  an  example,  consider  this  question : 


EXAMPLE 

Is  the  curve  y  =  x  log  x  concave  up- 
ward or  downward  at  i»  =  1  ? 

Solution.     Differentiating  twice, 

dy 


dx 

dry 
dx^' 


=  logx  + 1, 


1 

'  X 


d'y 


Hence  when  x  =  l,  -~  =  1  =  a  positive  number. 

(a/X'^ 


Hence  the 


curve  is  concave  upward  at  a^  =  1. 

dy~ 


Erom  the  first  derivative, 


dx 


:  log  1  +  1  =  1.     Also  from 


the  equation  of  the  curve,  when  x  =  1, 
y  =  log  1  =  0.  Arranged  in  a  table,  we 
have  the  results  set  down. 

The  appearance  of  the  curve,  in  the  neigh- 
borhood of  x  =  l,  is  now  determined,  and  we  may  construct  a 
small  arc  of  the  curve  at  that  point,  as  in  the  figure. 


X 

y 

dy 
dx 

dx-' 

1 

0 

1 

+ 

MAXIMA   AND   MINIMA  161 

PROBLEMS 

1.  Determine  the  direction  of  curvature  (concave  upward 
or  downward)  of  the  following  curves  at  the  points  indicated. 
Draw  a  figure  in  each  case. 

,    ^  .  TT      3  TT      5  TT 

(a)  \j=^\iix,  ^  =  1^  ~^^  -2' 

(h)  y  =  3x-a^,       x  =  l,2,  -3. 

(c)  y  =  e-%  x  =  0,  -1,2. 

(d)  y  =  xe'"",  x  =  0,  3. 

(e)  y  =  e"*  cos  x,      x  =  0. 
(/)  y  =  log{l  +  x),x  =  0,2. 

(g)y  =  i±^^  .^  =  1,-1,2.  ' 

x 

2.  Show  that  each  of  the  following  curves  is  always  concave 
upward  or  downward. 

(a)  yz=alogx,  (b)  y  =  Ax  —  x^.  (c)  y  =  ae~''. 

65.  Second  test  for  maxima  and  minima.  At  a  high  point, 
the  graph  is  concave  downward;  at  a  low  point,  concave 
upward.  The  directions  on  p.  152  may  therefore  be  stated 
in  another  form,  the  first  and  second  steps,  however,  remain- 
ing unchanged. 

Determination  of  maxima  and  minima  —  second  method. 

First  step.     Same  as  on  p.  152. 

Second  step.     Same  as  on  p.  152. 

Third  step.  Consider  any  critical  value  not  excluded  by  the 
conditions  of  the  problem,  and  substitute  this  in  the  second  deriva- 
tive of  the  function.  If  the  residt  is  negative,  the  function  has  a 
maximum  value  ;  if  positive,  a  minimum  value ;  if  zero,  the 
method  of  p.  152  must  be  resorted  to. 


162  ELEMENTARY   ANALYSIS 

Clearly,  this  second  method  should  be  used  only  if  the  second 
derivative  is  easily  obtained.  For  example,  the  method  of 
p.  152  is  preferable  for  the  problem  solved  on  p.  152.  On  the 
other  hand,  if  we  turn  to  the  problem  on  p.  94,  equation  (2), 

X 

the  differentiation  is  simple,  namely, 

d?M  ^^     864 

We  may  conveniently  test  the  critical  value 

x=:6  ( =  0  for  x  =  6 

\dx 

d?M~ 

by  substitution  in  the  second  derivative.     This  gives  — — 

dx-- 

=  a  positive  number.     Hence  Jf  is  a  minimum  when  x=  6. 

In  each  of  the  following  problems  the    function  is  easily 

differentiated  and  the  second  method  should  be  adopted. 

PROBLEMS 

1.  A  circular  sector  has  a  given  perimeter.  Show  that  the 
area  is  a  maximum  when  the  angle  of  the  sector  is  2  radians. 
(Area  of  a  sector  equals  ^  arc  times  radius.) 

2.  A  Gothic  window  consists  of  a  rectangle  surmounted  by 
an  equilateral  triangle.  The  perimeter  is  given.  What  must 
be  the  proportions  in  order  to  admit  as  much  light  as  possible  ? 

3.  Apply  the  second  test  for  maxima  and  minima  to 
problems  3  (a),  4  (a),  5,  6,  7  (a),  9,  10,  12,  14,  15,  and  19  of 
Art.  62. 

66.  Points  of  inflection.  The  problems  on  p.  110  illustrate 
the  aid  afforded  by  the  use  of  the  first  derivative  in  obtaining 


MAXIMA  AND  MINIMA  163 

an  accurate  plot.  The  second  derivative  is  also  of  service  in 
this  connection.  For  if  we  know  the  sign  of  the  second  de- 
rivative at  any  point  on  the  curve,  we  then  know  if  the  curve 
at  that  point  is  concave  upward  or  concave  downward  (p.  159), 
and  we  therefore  know  if  the  curve  at  that  point  lies  above  or 
below  the  tangent. 

Suppose  we  have  a  continuous  curve 
(as  in  this  figure)  such  that  in  passing 
along  it  from  A  to  (7,  the  curve  changes 
from  concave  upwards  to  concave  down- 
wards. Then  a  point  B  must  exist  such 
that 

to  the  left  of  B,  the  curve  is  concave  upwards; 

to  the  right  of  B,  the  curve  is  concave  downwards. 

The  point  B  is  called  a  point  of  inflection. 

The  second  derivative  (f"(x)  or  -^  j  is  positive  2it  edich  point 

of  the  arc  AB,  and  negative  at  each  point  of  the  arc  BC. 
Hence  at  B  the  second  derivative  is  zero. 

(J5)  At  points  of  inflectioD,  f"(x)  or  ^^  equals  zero. 

Solving  the  equation  resulting  from  (B)  gives  the  abscissas 
of  the  points  of  inflection.  To  determine  the  direction  of 
curving  in  the  vicinity  of  a  point  of  inflection,  testf"(x)  for 
values  of  x,  first  a  trifle  less  and  then  a  trifle  greater  than  the 
abscissa  at  that  point. 

Iif"(x)  changes  sign,  we  have  a  point  of  inflection,  and  the 
signs  obtained  determine  if  the  curve  is  concave  upwards  or 
concave  downwards  in  the  neighborhood  of  each  point. 

The  student  should  observe  that  near  a  point  where  the 
curve  is  concave  upwards  (as  at  A)  the  curve  lies  above  the 
tangent,  and  at  a  point  where  the  curve  is  concave  downwards 
(as  at  C)  the  curve  lies  below  the  tangent.  At  a  point  of  in- 
flection (as  at  B)  the  tangent  evidently  crosses  the  curve. 


164 


ELEMENTARY  ANALYSIS 


PROBLEMS 

Examine  the  following  curves  and  graphs  for  horizontal  tan- 
gents, points  of  inflection,  and  direction  of  curving. 

1.   y  =  3x^  —  4:X^  +  l. 

Solution.    f{x)  =  3x^-4.0^  +  1. 

Hence 

f(x)  =  12c^-12x'  =  12x\x-l). 
.*.  f'(x)  =  0  when  x  =  0,  ov  x  =  1. 
Differentiating  again, 

f\x)  =  S6x'-24.x. 
Using  (B), 

S6x'-24:x  =  0. 


x  =  i  and  x  =  0. 


Factoring, 


f\x)  =  36x{x-^). 


When  x<0,  f"  {x)=+',  and  when  x>0,  /' {x)=-. 

.*.  the  curve  is  concave  upwards  to  the  left  and  concave 
downwards  to  the  right  of  x  =  0  (A  in  figure). 

When  X  <  |,  /"  (x)  =  —-  and  when  x>^,f"  (x)  =  + . 

.'.  the  curve  is  concave  downwards  to  the  left  and  concave 
upwards  to  the  right  ol  x  =  ^  (B  in  figure). 

The  curve  is  evidently  concave  upwards  everywhere  to  the 
left  of  A,  concave  downwards  between  A  (0,  1)  and  B  (|,  ^), 
and  concave  upwards  everywhere  to  the  right  of  B. 

In  work  of  this  kind  it  is  well  to  tabulate  the  results  which 
aiford  a  check  on  the  plot.  We  follow  the  plot  from  left  to 
h^ght,  and  choose  for  the  initial  value  of  a;  a  value  to  the  left  of 
fT  the  least  root  of  f'(x)  =  0  and  f"(x)  =  0.  The  table  should  in- 
clude all  critical  values  of  x  (f'(x)  =  0)  and  also  those  for  which 
/"(aj)  =  0,  that  is,  values  determining  the  points  of  inflection. 
Furthermore,  intermediate  values  of  x  must  be  included,  as  in 
the  following  table,  which  the  student  should  carefully  study. 


MAXIMA  AND  MINIMA 


165 


X 

y 

d'y 

Eemarks 

~1 

0 

i 

1 

1 

2 

8 

1 

0 
17 

24 
0 

-¥ 

0 

48 

+ 
0 

0 

+ 
+ 

Concave  upward 
f  Point  of  inflection 
[  Horizontal  tangent 

Concave  downward 
1  Point  of  inflection 
[  Slope  of  tangent  =  —  ^^- 

Minimum  value  of  y 

Concave  upward 

2.    y-- 


3.    y  =  5  —  2x  —  oi:^. 


4.    y  =  0/^. 


Ans.  Concave    upwards    every- 
where. 

Ans.  Concave  downwards  every- 
where. 


Ans.  Concave  downwards  to  the 
left  and  concave  upwards 
to  the  right  of  (0,  0). 

Ans.  Concave  downwards  to  the 
left  and  concave  upwards 
to  the  right  of  (1,  -  2). 

Ans.  Concave  downwards  to  the 
left  and  concave  upwards 
to  the  right  of  (a,  b). 

Ans.  Concave  downwards  to  the 

left  and  concave  upwards 

4a 

'3 

8.    x^  —  3  hx^  +  a?y  =  0.  Ans.  Point      of     inflection      is 


5.    ?/  =  a^-3a^-9aj  +  9. 


6.    y=za-\-(x—  by 


7.   a-y  = ax^  -f  2  a^. 

^3 


to  the  right  of  [  a, 


9.   y  =  x^. 


'•f 


Ans.  Concave    upwards    every- 
where. 


166 


ELEMENTARY  ANALYSIS 


10.   Soc^  —  d  0:^  —  27  x  +  SO.    Ans,  x  =  —  l,  gives  max.  =  45 ; 

ives  mill.  =  —  51. 


x  =  3j  g 

11.   2x^-21x^  +  36x--20.  Ans.  x  =  l, 

x  =  6,g: 

Ans,  x  =  ly  g 
x  =  3,g: 

13.  2a^-15x^  +  36x  +  10.  Ans.  x  =  2,  g 

x.=  3,  g 

14.  a^-9x'  +  15x-3. 


12..  t-2x^  +  3x-\-l. 
3 


15.  a^-3x^  +  6x-it-10. 

16.  o(^  —  5x^-\-5x^-{-l. 


Ans.  x  =  l,  g 

'-      ^  =  5,  g- 
Ans.  No  max.  or  min. 


ves  max.  =  —  3  ; 
ves  min.  =  — 128. 

ves  max.  =|-; 
ves  min.  =  1. 

ves  max.  =  38  ; 
ves  min.  =  37. 


ves  max.  =  4 ; 
ves  min.  =  —  28. 


Ans.  x  =  l,  gives  max.  =  2 ; 

x  =  3,  gives  min.  =  —  26; 
x  =  Oj  gives  neither. 


CHAPTER   XI 

RATES 

67.   Velocity  and  ,  p  q 

acceleration.      Con-  oj  x  '        Xx       ^         x 

sider  a  moving  point 

on  the  a;-axis.     Its  distance  from  the  origin  (=  x)  is  a  function 
of  the  time.     That  is,  symbolically, 

(1)  x=f{t). 

Suppose  that  P  is  the  position  when  the  time  is  t  seconds, 
and  suppose,  further,  that  the  elapsed  time  from  P  to  Q  is  A^ 
seconds.     Also  let  PQ  =  Ax,     Then  the  quotient 

Ax 

(2)  —  =  average  velocity  of  the  motion  from  P  to  Q. 

Ax 
The  velocity  at  P  is  the  limit  of  the  value  of  —  when  Q  is 

taken  nearer  and  nearer  to  Q ;  that  is 

(3)  f =irJot=-'-"''^*^ 

Considered  without  the  idea  of  motion,  equation  (1)  asserts 
that  the  variable  x  changes  as  t  changes.     Then  we  say 

(4)  —  =  time  rate  of  change  of  x. 

Similarly,  if  y  and  z  are  functions  of  the  time,  then  -^^  and  — 
are  the  time  rates  of  change  of  y  and  z  respectively. 

The  point  P  may  move  from  the  point  0  with  constant 

167 


168  ELEMENTARY  ANALYSIS 

speed  Vq.      Then  the  distance  moved  is  the  product  of  -the 
speed  (or  velocity)  and  the  time.     That  is,  equation  (1)  now  is 

(5)  X  =  VqL 

,   If,  however,  the  velocity  changes,  the  time  rate  of  change  of 
velocity  is  called  the  acceleration^  and  equation  (4)  asserts  that 

(6")  acceleration  =  -—  • 

^  ^  dt 

From  (3)  we  may  also  write, 

/r7\  ^  j_'  f^^  a  OS 

(7)  acceleration  =  —  =  — -  • 

^  ^  dt       dt^ 

Similarly,  any  derivative  may  be  interpreted  as  a  rate  of 

change.     For  example,  ^  =  rate  of  change  of  y  with  respect 

dx 
to  i;);,  if  2/  is  a  function  of  x. 
\  The  case  frequently  arises  when  in  the  equation 

both  X  and  y  are  functions  of  the  time.    If  the  time  rate  of  change 

in  X  I  =  —  )  is  known,  then  —  can  he  found. 
\     dt)  '  dt  ^ 

PROBLEMS 

1.  A  particle  slides  along  the  curve,  y'^  —  ^x  =  0,  so  that  it 
moves  in  the  XX^  direction  at  the  constant  rate  of  3  ft.  a 
second.  How  fast  is  it  moving  in  the  FF'  direction  (a)  at 
any  point  {x,  y)  ;  (h)  at  the  point  (4,  4)  ?  # 

Solution.     Given  —  =  3,  required  -^  • 
dt  dt 

As  y  and  x  in  the  equation  y'^  —  4.x  =  0  are  functions  of  t, 
we  differentiate  the  equation  with  respect  to  t.  This  gives 
(p.  144), 

m      dt 


RATES  169 

Solving  for  — , 

(a)  ^  =  ?  ^  =  ?ft.  a  second ;  (6)  ^  =  ^  =  ?  ft.  a  second. 
^  ^   dt      y  dt      y  '  ^  ^  d^      4     2 

2.    Find  the  rate  of  change  of  the  volume  and  of  the  surface 

of  a  sphere  with  respect  to  the  radius. 

dv 
Solution.     Let  y  be  the  volume  and  x  the  radius.    Required  —• 

cccc 

Now  yz=-7rci^,  .-.  -^  =  4  7ra^, 

3  dx 


which  varies^as  the  square  of  the  radius.     The  rate  of  change 

of  the  surface  z  with  respect  to  the  radius  x  is  found  from  the 

•  ^^ 

relation  z  =  4:  ttx^  to  be  —  =8  ttx,  and  it  varies  as  the  first 

dx 

power  of  the  radius. 

3.  An  express  train  and  a  balloon  start  at  points  12  mi. 
apart.  The  former  runs  50  mi.  per  hour,  and  the  balloon  rises 
vertically  at  the  rate  of  8  mi.  per  hour.  How  fast  are  they 
separating  at  the  end  of  10  min.  ? 

Solution.     Let  the  figure  represent  the  state  of  affairs  at  the 
expiration  of  any  number  (=^)  of      d 
minutes.     That  is, 

y 

BD  is  the  path  of  the  balloon, 
AC  \s  the  path  of  the  train,  l^ 

and  y  =  number  of  miles  traveled  by  balloon, 

0.'  =  number  of  miles  traveled  by  train, 
z  =  distance  apart. 

Evidently,  x,  y,  and  z  are  all  functions  of  the  time. 
Now  get  the  relation  between  these  variables. 
Evidently, 

(1)  z''  =  f+(x  +  i2y. 

Differentiate  with  respect  to  t.     This  gives 

2z  —  =  2y^-\-2(x  +  12)  —  ' 
dt        '^  dt        ^  ^  dt 


170  ELEMENTARY   ANALYSIS 

^  ^  dt     y  dt     ^  ^  dt 

\      This  result  gives  the   velocity  of   separation  at   any  time, 
'  The  prc^lem  calls  for  this  velocity  after  10  min.     Hence 

x  =  S^  miles,  y  =  U  miles,  —  =  50,  ^  =8. 
/  dt  \dt 

We  find  from  (1)  the  value  of  2;  =  20.4  mi.     Substituting  in 

(2)  we  find  --  =  50.3  mi.     Ans. 
^  ^  dt 

This  problem  shows  that  in  many  cases  the  following  rule 
should  be  followed. 

1.  Draw  a  diagram  to  represent  the  state  of  affairs  at  any 
instant  {after  t  seconds). 

2.  In  this  diagram  mark  the  variable  elements  with  x,  y,  z,  etc, 

3.  Find,  the  equation  connecting  x,  y,  z,  etc. 

4.  Differentiate  this  result  with  respect  to  t. 

5.  Substitute  the  given  data  in  this  result. 

4.  The  radius  of  a  soap  bubble  is  increasing  at  the  rate  of 
2  in.  a  second.  How  fast  is  the  volume  increasing,  (a)  at  any 
time,  (5)  when  the  radius  is  3  in.  ? 

Ans.  (a)  8  vrr^  cu.  in.  a  second,  (b)  72  tt  cu.  in.  a  second. 

5.  Solve  the  problem  similar  to  the  preceding,  where  vol- 
ume is  replaced  by  surface. 

6.  At  what  point  of  Ex.  1  are  the  ordinate  and  the  abscissa 
,  increasing  at  the  same  rate  ?  Aiis.  (1,  2). 

\7.    Where  in  the  first  quadrant  does  the  arc  increase  twice 
as  fast  as  the  sine  ?  Ans.  60°. 

8.  A  man  6  ft.  tall  walks  away  from  a  lamp  post  10  ft. 
high  at  the  constant  rate  of  4  mi.  an  hour.  How  fast  does 
the  shadow  of  his  head  move  ?  (Use  similar  triangles  to  de- 
termine the  relation  between  the  man's  distance  from  the 
lamp  post  and  the  distance  of  the  shadow.) 

Ans.  10  mi.  an  hour. 


/ 


RATES  171 

^^9.  Find  the  rate  of  change  of  the  area  of  a  square,  when 
the  side  x  is  increasing  at  the  rate  of  k  in.  a  second. 

A71S.  2  kx  sq.  in.  a  second. 

/^lO.   A  ladder   25  ft.  long   rests  against   a   house.     A  man 

takes  hold  of  the  lower  end  and  moves  it  aw^ay  at  the  rate  of  2 

ft.  a  second.     How  fast  is  the  top  of  the  ladder  descending 

when  the  bottom  is  7  ft.  from  the  house  ?   Ans.  7  in.  a  second. 

C^yfi.  Two  particles  start  together  from  the  origin  along 
y^  —  9x  =  0  and  of  -{- f/  —  34:  x  =  0  respectively.  The  former 
has  a  speed  in  the  YY'  direction  of  3  ft.  a  second,  and  the 
latter  of  4  ft.  a  second.  Which  goes  through  the  point  (25, 15) 
with  the  greater  speed  in  the  XX'  direction  ? 

A71S.  One  on  the  parabola. 

12.  Find  how  fast  the  man's  shadow  in  problem  8  is 
lengthening.  Ans.  6  mi.  an  hour. 

P  13.    A  circular  plate  expands  by  heat  so  that  the  radius  in-   y 
creases  uniformly  at  the  rate  of  .01  in.  a  minute.     At  what 
rate  is  the  surface  increasing  when  the  radius  is  2  in.  ? 

Ans.  .879  sq.  in.  a  minute. 

14.  Water  runs  into  a  barrel  at  the  rate  of  2  cu.  ft.  a  min- 
ute, but  leaks  at  the  bottom  at  the  rate  of  1  cu.  ft.  a  minute. 
Assuming  the  barrel  to  be  a  right  cylinder  of  radius  1  ft.  and 
of  height  5  ft.,  how  long  will  it  be  before  water  runs  over  the 
top  ?  Ans.  7  min.  51  sec. 

/^15.  A  boy  starts  flying  a  kite.  If  it  moves  horizontally  at 
the  rate  of  2  ft.  a  second,  and  rises  at  the  rate  of  5  ft.  a  second, 
how  fast  is  the  string  being  paid  out  at  the  end  of  3  min.  ? 

Ans.  5.38  ft.  a  second. 

^^0^^.  At  what  point  of  the  curve  a??/  -f  25  =  0  v^ill  a  particle 
move  in  the  x  and  y  direction  at  the  same  rate?    Ans.  (—  5,  5). 

/^^JW^A  man  standing  on  a  dock  is  drawing  in  the  painter  of 
a  boat  at  the  rate  of  2  ft.  a  second.  His  hands  are  6  ft.  above 
the  bow  of  the  boat.  How  fast  is  the  boat  moving  when  it  is 
8  ft.  from  the  dock  ?  Ans.  f  ft.  a  second. 


> 


\ 

172  ELEMENTARY  ANALYSIS 

18.  The  volume  and  the  radius  of  a  cylindrical  boiler  are 
expanding  at  the  rate  of  1  cu.  ft.  and  .001  ft.  per  minute 
respectively.  How  fast  is  the  length  of  the  boiler  changing 
when  the  boiler  contains  60  cu.  ft.  and  has  a  radius  of  2  ft.  ? 

Ans.  0.78  ft.  a  minute. 


/: 


19.  An  equilateral  triangular  sheet  of  rubber  is  stretched  so 
that  it  always  keeps  its  shape,  but  expands  at  the  rate  of  3  sq. 
a  Qiinute.     How  fast  is  its  side  increasing  when  4  in.  long  ? 

Ans.   .866  in.  a  minute. 


20.  The  rays  of  the  sun  make  an  angle  of  30°  with  the  hori- 
zon. A  ball  is  thrown  vertically  upward  to  a  height  of  64  ft. 
How  fast  is  its  shadow  traveling  along  the  ground  just  be- 
fore the  ball  hits  the  ground?     (Use  s  =  ^gf  and  v  =  gt,) 

Ans.  110.8  ft.  a  second. 

21.  A  man  is  walking  over  a  bridge  at  the  rate  of  4  mi.  an 
hour,  and  a  motor  boat  passes  beneath  him  at  the  rate  of  8  mi. 
an  hour.  If  the  bridge  is  20  ft.  above  the  boat,  how  fast  are 
they  separating  3  min.  later?  Ans.  8.95  mi.  an  hour. 

22.  A  ship  is  anchored  in  18  ft.  of  water.  The  cable  passes 
over  a  sheave  on  the  bow  6  ft.  above  the  surface  of  the  water. 
If  the  cable  is  taken  in  at  the  rate  of  1  ft.  a  second,  how  fast 
is  the  ship  moving  when  there  are  30  ft.  of  cable  out  ? 

Ans.  -|  ft.  a  second. 

23.  A  boy  rows  out  for  a  swim  against  a  tide  running  in  at 
the  rate  of  i  mi.  an  hour  horizontally.  He  dives  off  and 
swims  parallel  to  the  coast  at  the  rate  of  2  mi.  an  hour.  How 
fast  are  he  and  his  boat  separating  at  the  end  of  15  min.  ? 

Ans.  2.06  mi.  an  hour. 

24.  Four  men  standing  5  ft.  from  a  house  are  hoisting  a 
piano  to  the  third-story  window  by  means  of  a  block  and 
tackle.  If  the  window  is  50  ft.  up,  and  the  men  pull  in  the 
rope  at  the  rate  of  10  ft.  a  minute  and  back  away  from  the 
building  at  the  rate  of  5  ft.  a  minute,  how  fast  is  the  piano  ris- 
ing at  the  end  of  the  first  minute  ?        Ans.  10.98  ft.  a  minute. 


KATES  173 

25.  Find  at  what  points  on  y  =  x^  —  4:  the  rate  of  increase 
of  y  with  respect  to  x  is  equal  to  the  rate  of  increase  of  x 
with  respect  to  y.  Ans.  (±  |-,  —  -V). 

26.  Assuming  the  volume  of  wood  in  a  tree  is  proportional 
to  the  cube  of  its  diameter  and  that  the  latter  increases  uni- 
formly year  by  year,  show  that  the  rate  of  growth  when  the 
diameter  is  3  ft.  is  36  times  as  great  as  when  it  is  6  in. 

27.  A  rectangular  sheet  of  metal  is  subjected  to  a  pressure 
which  expands  it  only  lengthwise  and  at  a  rate  of  2  in.  a  min- 
ute. Find  how  fast  the  area  of  the  sheet  is  enlarging,  if  it  is 
7  in.  wide.  Ans.  14  sq.  in.  a  minute. 

28.  Water  flows  from  a  faucet  into  a  hemispherical  basin 
of  diameter  14  in.  at  the  rate  of  2  cu.  in.  a  second.  How  fast 
is  the  water  rising  (a)  when  the  water  is  halfway  to  the  top, 
(b)  just  as  it  runs  over  ?  (The  volume  of  a  spherical  segment 
is  l7rr^^  +  |-7r/i^) 

A71S.  (a)  .013  in.  a  second ;  (b)  .005  in.  a  second. 

29.  Sand  is  being  poured  on  the  ground  from  the  orifice  of 
an  elevated  pipe,  and  forms  a  pile  which  has  always  the  shape 
of  a  right  circular  cone  whose  height  is  equal  to  the  radius  of 
the  base.  If  sand  is  falling  at  the  rate  of  6  cu.  ft.  per  second, 
how  fast  is  the  height  of  the  pile  increasing  when  the  height 
is  5  ft.?  Ans.   .076  ft.  per  second. 

30.  An  aeroplane  is  528  ft.  directly  above  an  automobile, 
and  starts  east  at  the  rate  of  20  mi.  an  hour  at  the  same  time 
that  the  automobile  starts  east  at  the  rate  of  40  mi.  an  hour. 
How  fast  are  they  separating  in  6  min.  ? 

Ans.  19.92  mi.  an  hour. 

31.  A  ship  is  41  mi.  due  north  of  a  second  ship.  The  first 
sails  south  at  the  rate  of  8  mi.  an  hour ;  the  second,  east,  at  the 
rate  of  10  mi.  an  hour.  How  rapidly  are  they  approaching 
each  other  ?     How  long  will  they  continue  to  approach  ? 

Ans.  For  2  hr. 


174  ELEMENTARY  ANALYSIS 

32.  A  railroad  train  is  running  along  a  curve  in  the  form  of 
2/2  =  500  X,  tHe  axis  of  the  parabola  being  east  and  west.  If 
the  train  is  going  due  east  at  the  rate  of  30  mi.  an  hour,  how 
fast  is  the  shadow  moving  along  the  wall  of  a  station  which 
runs  north  and  south,  when  the  train  is  500  ft.  east  of  the 
wall,  provided  the  sun  is  just  rising  in  the  east? 

Ans.  15  mi.  an  Jiour. 

33.  One  side  of  a  rectangle  inscribed  in  a  circle  of  radius 

5  cm.  expands  at  the  rate  of  2  cm.  a  minute.  Find  (a)  how 
farSt  the  area  of  the  rectangle  is  changing  when  the  above  side 
is  8  cm. ;  {h)  how  long  the  above  side  is  when  the  area  is  not 
changing.     (Get  the  area  as  a  function  of  the  side.) 

Ans.  (a)  —  9J  cm.  a  minute;  (h)  3.5  cm.  a  minute. 

34.  Work  out  similar  problems,  for  cylinders  and  cones  in- 
scribed in  a  fixed  sphere,  where  the  radii  of  their  bases  vary. 

35.  A  revolving  light  sending  out  a  bundle  of  parallel  rays 
is  at  a  distance  of  |-  a  mile  from  the  shore  and  makes  1  revo- 
lution a  minute.  Find  how  fast  the  light  is  traveling  along 
the  beach  when  at  a  distance  of  1  mi.  from  the  nearest  point 
of  the  beach.  Ans,  15.7  mi.  per  minute. 

36.  A  kite  is  150  ft.  high,  and  200  ft.  of  string  are  out.  If 
the  kite  starts  drifting  horizontally  and  away  from  the  flyer, 
at  the  rate  of  4  mi.  an  hour,  how  fast  is  string  being  paid  out  ? 

Ans.  2.64  mi.  an  hour. 

37.  A  solution  is  poured  into  a  conical  filter  of  base  radius 

6  cm.  and  height  24  cm.  at  the  rate  of  2  cu.  cm.  a  second, 
Avhich  filters  out  at  the  rate  of  1  cu.  cm.  a  second.  How  fast 
is  the  level  of  the  solution  rising  when  (a)  one  third  of  the 
way  up,  (h)  at  the  top  ?  Ans.  (a)  .079  cm.  a  second, 

(b)  .009  cm.  a  second. 


CHAPTER   XII 

DIFFERENTIALS 

68.   We  have  already  emphasized  that  the  symbol  4 

dy 
dx 

was  to  be  considered  not  as  an  ordinary  fraction  with  dy  as 
numerator  and  dx  as  denominator,  but  as  a  single  symbol  de- 
noting the  limit  of  the  quotient 

Ao? 

as  ^x  approaches  the  limit  zero. 

Problems  do  occur,  however,  where  it  is  convenient  to  give 
meanings  to  dx  and  dy  separately.  How  this  may  be  done  is 
explained  in  what  follows. 

Definition,  li  f  (x)  is  the  derivative  of  f(x)j  and  ^x  is  an 
arbitrarily  chosen  increment  of  x,  then  the  differential  of  f{x), 
denoted  by  the  symbol  df(x),  is  defined  by  the  equation 

(A)  df(x)=f(x)Ax, 

If  now  f(x)  =  X,  then  /'  (x)  =  1,  and  (A)  reduces  to 

dx  =  Ax, 

Hence,  when  x  is  the  independent  variable,  the  differential  of 
x(=dx)  is  identical  with  Ax.  If  y=f(x),  (A)  may  be  written 
in  the  form 

(B;  dy=f(x)dx.* 

*  On  account  of  the  position  which  the  derivative  fix)  here  occupies,  it  is 
sometimes  called  the  differential  coefficient. 

'  175 


176 


ELEMENTARY   ANALYSIS 


Y 

V 

1 

dy 

y 

-f^dx-'S 

y 

-^^ 

o 

y 

M    J 

M'X 

The  differential  of  a  function  equals  its  derivative  multiplied 
by  the  differential  of  the  independent  variable. 

Let  us  illustrate  what  this  means  geo- 
metrically. 

Let  /'  (x)  be  the  derivative  of  y  =f(x) 
at  P. 

Take  dx  =  PQ,  then 

dy  =f'  (x)  dx  =  tan  r  •  PQ 

=  f^.-PQ=QT. 

Therefore  dy,  or  df(x)j  is  the  increment  (=  QT)  of  the  ordi- 
nate of  the  tangent  corresponding  to  dx. 

This  gives  the  following  interpretation  of  the  derivative  as 
a  fraction : 

If  an  arbitrarily  chosen  increment  of  the  independent  variable 
xfor  a  point  P{x,  y)  on  the  curve  y  =f(x)  be  denoted  by  dx,  then 
in  the  derivative 

dy 


dx 


-f  (x)  =  tan  T 


dy  denotes  the  corresponding  increment  of  the  ordinate  of  the  tan- 
gent to  the  curve  at  the  point  P. 

The  student  should  observe  that,  while  the  differential  and 
the  increment  of  the  independent  variable  are  always  equal 
{dx  =  IS.x),  the  same  is  not  true  of  the  dependent  variable.  In 
the  figure,  A?/  =  increment  of  2/  =  QP\  but  dy  =  differential  of 
y=QT. 

69.  Formulas  for  finding  differentials.  Since  the  differen- 
tial of  a  function  is  its  derivative  multiplied  by  the  differential 
of  the  independent  variable,  it  follows  at  once  that  the  formu- 
las for  finding  differentials  may  be  obtained  from  those  for 
finding  derivatives  (Art.  42)  by  multiplying  by  dx. 

This  gives 

I  d(c)  =  0. 

II  d  (x)  =  dx. 


Ill 

d(io  +  v  —  w)  =du  +  dv  —  dw. 

IV 

d  (cv)  =  cdv. 

V 

d(uv)=  udv  +  vdu. 

VI 

d(y")  =  nv''~^dv. 

via 

d  («")  =  w«"-i  dx. 

VII 

jfu\     vdu  —  udv 

Vila 

\cj       c 

VIII 

d(\og^v)  =  log^e^. 

IX 

d  (a")  =  a"  log  a  dv. 

IX  a 

d  (e")  =  e"  dv. 

X 

d(sinv)=  cosvdv. 

XI 

d  (cos  v)—  —  sin  v  dv. 

XII 

d  (tan  v)  =  sec-  v  dv,  etc. 

XVI 

d  fare  sin  «)= — ,  etc. 

177 


Vl-v' 


The  term  differentiation  also  includes  the  operation  of  find- 
ing differentials. 

In  finding  differentials,  the  easiest  way  is  to  find  the  deriva- 
tive as  usual,  and  then  multiply  the  result  by  dx. 

:e:xAMPLEs 
1.   Find  the  differential  of 

x-{-3 

Solution. 

^        .fx-\-3\^(x'-\-'S)d{x^3)-{x^-3)d{x'  +  3) 
^        \x'  +  3y  (x"  +  3)2 

^  {x'-{-3)dx-{x-\-3)2xdx  ^  (3  -  6.-^  -  x") dx        . 

{x^  +  3f  {x'  +  Sf        *        ^^' 


178  ELEMENTARY   ANALYSIS 

2.  Find  dy  from 

Solution. 

2Wxdx-2a^ydy=-^. 

,'.  dy  =  ---dx.    Ayis, 
o?y 

3.  Find  dp  from 

9" 
Solution.  2  pclp  =  —  a^  sin  2  ^  .  2  dO. 

,            a?  sin  2  6  -.^ 
.  • .  dp  = dd, 

9 

4.  Find  d  [arc  sin (3^-4  f)']. 

Solution. 

d[arcsin(3^-4f^)]^^^^~^^'^       =     ^^^     .     Ans. 

Vl-(3^-4^/      Vl-^' 

PROBLEMS 

Differentiate  the  following,  using  differentials. 

1.  yz=za^  —  hx'^+cx  +  d.  dy  =  (S  ax^  —  2  bx  +  c)dx, 

2.  y=  2x^—^  x^  +  6aj-H5.       dy={5x^  —  2xr^  —  6  a;-^)^^!. 

3.  2/  =  (a^  —  i^)^  c?2/  =  —  10  x{ar  —  a^^^c^aj. 


4.    ?/  =  Vl  +  aJ^.  (^V  =  — =^=zz  dx. 

^       Vl+a;2 

s-  y—-7z — ^*  '              ^y= 7, — .dx. 

{i-^-x'y  ^    (i  +  x''y+'^ 

6.  ^  =  logVl-.?.  ^^  =  2(^ir3r)- 

7.  ?/  =  (e^  4-  e-'^y.  dy  =  2(e2^  -  e-^'')dx. 

8.  y  —  e^  log  ^, 


d^  =  e'l  log  cj;  +  -  Ic^ic. 


9.    8  =  ^-"-  ~^     .  ds=#(^^?^ ^'Vd^. 


DIFFERENTIALS  179 

10.  p  =  tan  <^  +  sec  <^.  c?p  =  i-±^i5_^  dd). 

11.  r  =  i  tan^(9  +  tan  (9.  dr  =  seG^edO, 

12.  f(x)  =  (log  xy,  f(x)dx  =  ^(M^^ . 

X 
13.      cl>(t)= ^3.  ^'(^C?^:  ^^'^^ 


70.  Infinitesimals.  An  infinitesimal  is  a  variable  whose 
value  decreases  numerically  and  approaches  zero  as  a  limit. 

In  Art.  68  it  has  been  shown  that  the  differential  and  incre- 
ment of  the  independent  variable  are  identical.  Equation  (B) 
of  that  section  defines  the  differential  of  the  dependent  variable. 
Clearly,  if  dx  is  an  infinitesimal,  then  also  dy  is  an  infinitesimal. 
In  the  Integral  Calculus,  we  have  to  do  with  expressions  of  the 
form  cj)(x)dx,  called  ^*  differential  expressions.^'  Then,  if  dx  is 
an  infinitesimal,  so  also  is  the  product  <t>(x)dx. 


CHAPTER   XIII 

INTEGRATION.  RULES  FOR  INTEGRATING  STANDARD 
ELEMENTARY  FORMS 

71.  Integration.  The  student  is  already  familiar  with  the 
mutually  inverse  operations  of  addition  and  subtraction,  multi- 
plication and  division,  involution  and  evolution.  From  the 
Differential  Calculus  we  have  learned  how  to  calculate  the 
derivative  f{x)  of  a  given  function  f{x),  an  operation  indi- 
cated by 

or,  if  we  are  using  differentials,  by 

df(x)=f\x)dx. 

The  problems  of  the  Integral  Calculus  depend  on  the  inverse 
operation ;  namely, 

To  find  a  function  f(x)  vjJiose  derivative 

(A)  f(x)=<f>{x) 

is  given. 

Or,  since  it  is  customary  to  use  differentials  in  the  Inte- 
gral Calculus,  we  may  write 

(B)  df(x)  =f(x)  dx  =  <i>  (x)  dx, 

and  state  the  problem  as  follows : 

Having  given  the  differential  of  a  function,  to  find  the  function 
itself 

The  function  f(x)  thus  found  is  called  an  integral  of  the  given 
differential  expression,  the  process  of  finding  it  is  called  inte- 

180 


INTEGRATION  181 

gration,  and  the  operation  is  indicated  by  writing  the  integral 
sign  ^    j    in  front  of  the  given  differential  expression.     Thus, 

(0)  jf\x)dx=f{x), 

read,  an  integral  of  f'(x)dx  equals  f{x).     The  differential  dx 
indicates  that  x  is  the  variable  of  integration.     For  example, 

(a)  If  f{x)  =  x?,  then  /'(^)  dx  =  3  Qi?dx, 
and  I  3  x'^dx  =  ^x^, 

(h)  If  fix)  =  sin  X,  then  /'  (x)  dx  =  cos  xdXy 

and  I  cos  xdx  =  sin  x, 

(c)  If  f(x)  =  arc  tan  x,  then  /'(a.*)  =  — — - , 


and  I =  arc  tan  x. 

J  1+x^ 

Let  us  now  emphasize  what  is  apparent  from  the  preceding 
explanations,  namely,  that 

Differentiation  and  integration  are  inverse  operations. 
Differentiating  (C)  gives 

(D)  d  Cf'(x)  dx  =  fix)  dx. 

Substituting  the  value  oif{x)dx\_=df{x)']  from  (B)  in  (O), 
we  get 

{E)  fdf(x)=f(x). 

Therefore,  considered  as  symbols  of  operation,  —  and  I  -^  dx 

dx        J 

*  Historically  this  sign  is  a  distorted  »S,  the  initial  letter  of  the  word  smn. 
Instead  of  defining  integration  as  the  inverse  of  differentiation  we  may  define 
it  as  a  process  of  summation,  a  very  important  notion  which  we  consider  in 
Chapter  XVI. 


182  ELEMENTARY  ANALYSIS 

are  inverse  to  each  other ;  or,  if  we  are  using  differentials,  d  and 
I   are  inverse  to  each  other. 

When  d  is  followed  by  I  they  annul  each  other,  as  in  (Z)), 
but  when  I  is  followed  by  d,  as  in  (E),  that  will  not  in  gen- 
eral be  the  case  unless  we  ignore  the  constant  of  integration. 
The  reason  for  this  will  appear  at  once  from  the  definition  of 
the  constant  of  integration  given  in  the  next  section. 

72.  Constant  of  integration.  Indefinite  integral  From 
the  preceding  section  it  follows  that 

since  d  (ar^)  —  3  xHx,  we  have    I  3  :)(?dx  =  x^ ; 
since  d(x^+2)  =  ^ x^dx,  we  have   I  3 xHx  =  a^  +  2 ; 

since  d{oi?  —  l)  =  ^  xHx,  we  have   I  3 x^dx  =  x^  —  7. 

In  fact,  since  d(x^+  0)=3 a^dXy 

where  C  is  any  arbitrary  constant,  we  have 

C3a^dx  =  a^+a 

A  constant  O  arising  in  this  way  is  called  a  constant  of 
integration.'^  Since  we  can  give  C  as  many  values  as  we 
please,  it  follows  that  if  a  given  differential  expression  has 
one  integral,  it  has  infinitely  many  differing  only  by  constants. 
Hence 

]f(x)dx=f(x)+C', 


s> 


and  since  C  is  unknown  and  indefinite^  the  expression 

is  called  the  indefinite  integral  off'(x)dx. 

*  Constant  here  means  that  it  is  independent  of  the  variable  of  integration. 


INTEGRATION  183 

It  is  evident  that  if  <^  (x)  is  a  function  the  derivative  of  which 
is  f(x),  then  <l>{x)-\-  C,  where  C  is  any  constant  whatever,  is 
likewise  a  function  the  derivative  of  which  isf(x).    Hence  the 

Theorem.  If  two  functions  differ  by  a  constant,  they  have  the 
same  derivative. 

73.  Rules  for   integrating   standard  elementary  forms. 

The  Differential  Calculus  furnished  us  with  a  General  Rule  for 
Differentiation  (p.  104).  The  Integral  Calculus  gives  us  no 
corresponding  general  rule  that  can  be  readily  applied  in 
practice  for  performing  the  inverse  operation  of  integration. 

Each  case  requires  special  treatment,  and  we  arrive  at  the 
integral  of  a  given  differential  expression  through  our  previous 
knowledge  of  the  known  results  of  differentiation. 

Integration  then  is  essentially  a  tentative  process,  and  to 
expedite  the  work,  tables  of  known  integrals  are  formed  called 
standard  forms.  To  effect  any  integration  we  compare  the 
given  differential  expression  with  these  forms,  and  if  it  is 
found  to  be  identical  with  one  of  them,  the  integral  is  known. 
If  it  is  not  identical  with  one  of  them,  we  strive  to  reduce  it 
to  one  of  the  standard  forms  by  various  methods,  many  of 
which  employ  artifices  which  can  be  suggested  by  practice 
only. 

From  any  result  of  differentiation  may  always  be  derived  a 
formula  for  integration. 

The  following  two  rules  are  useful  in  reducing  differential 
expressions  to  standard  forms. 

(a)  The  integral  of  any  algebraic  sum  of  differential  expressions 
equals  the  same  algebraic  sum  of  the  integrals  of  these  expressions 
taken  separately. 

Proof     Differentiating  the  expression 


j  dit  +  j  d'y  —  j  div, 


u,  v,  w  being  functions  of  a  single  variable,  we  get 

du  +  dv  —  div.  (Ill,  p.  113) 


184  ELEMENTARY  ANALYSIS 

[1]  .',  I  {du  +  dv —  dw)=  jdu+  I  dv  —  j  div. 

(b)  A  constant  factor  may  he  written  either  before  or  after  the 
integral  sign. 
Proof     Differentiating  the  expression 

a  i  dv 
gives  adv.  (IV,  p.  113) 

[2]  /.  j  adv  =  a\  dv. 

On  account  of  their  importance  we  shall  write  the  above  two 
rules  as  formulas  at  the  head  of  the  following  list  of 

STANDARD   ELEMENTARY   FORMS 
[1]  (  {du  -\-dv  -  dw)  =  \du  +  \dv  —  (dw. 

[2]  \adv  =  a  \dv. 

[3]  (doc  =  oc^€. 

[4]  (v-dv=^^^^-C,  n^-1 

[5]        r^-io^v+c 

J   V 

=  log  V  +  log  c  =  log  ev. 

[Placing  C  —  log  c] 

[6]  Ca-dv=-^^^C. 

•/  log  a 

[7]  .(e'^dv  =  e''+  C. 
[8]  i^invdv  =-  cos  v  -\-  C. 

[9]  Icos  V  dv  =  sin  v -^  C. 

[10]  t  sec^  V  dv  =  tan  v  -i-  C. 


mXEGRATION  185 

[11]  (eo^ed^vdv  =  — cotv -h  C. 

[12]  i^ecvtanvdv  =  ^ecv  +  C. 

[13]  icsaveotvdv  =— cfiev -\- C. 

[14]  Ttan  vdv  =  log  sec  v  +  C, 

[15]  (  cot  i;  dv  =  log  sin  v  -\-  C. 

[16]  r    ^^   ^=:iarctan-+C 

[17]  r      ^^       =  arc  sin  -  +  C. 

Proo/o/[3].     Since 

c^(a^+(7)  =  da^,  (II,  p.  113) 

then  i  dx  =  X  +  C. 

Proof  of  [4].     Since 

d  fj^\  =  v-dv,  (YII,  p.  113) 

\7i  +  ly 

n  +  1 

This  holds  true  for  all  values  of   n  except  n  =  —  1.     For, 
when  n  =  —  l,  [4]  gives 

Jv-hlv  =-J^^  4-  C=^  +  C=  00  4-  O, 

which  has  no  meaning. 

The  case  when  n  =  —  1  comes  under  [5]. 
Proof  of  [^5'].     Since 

d  (log  ^  +  C)  =  — ,  (YIII  a,  p.  113) 

we  get  I  ^ —  =  log  V  -{-  C. 

J    V 


186  ELEMENTARY   ANALYSIS 

The  results  we  get  from  [5]  may  be  put  in  more  compact 
form  if  we  denote  the  constant  of  integration  by  log  c.     Thus, 

=log  V  +  log  C  =  log  CO. 

V 

Formula  [5]  states  that  if  the  expression  under  the  integral 
sign  is  a  fraction  whose  mimerator  is  the  differential  of  the  de- 
nominator,  then   the   integral   is  the  natural   logarithm   of  the 

denominator. 

PROBLEMS 

For  formulas  [l]-[5]. 

Verify  the  following  integrations. 

x^dx  =  — h  C,  by  [4],  where  v-=^x  and  ri  =  6. 

2.  j  aQ(?dx  =  a\  Mx  (by  [2]) 

=  ^+C.  (by  [4]) 

4 

3.  C{2^-bx'-3x  +  4.)dx 

=  C2  xhlx  -  ("5  xHx  -  Cs  xdx  +  A  dx  (by  [1]) 

=  2  Coi^dx  -  5  Cx'^dx  -  3  Cxdx  +  4  Cdx     (by  [2] ) 

x'     5x'     Sx\    .      ,  ^      • 
=  2-T— 2-  +  '"+^-    ' 

Note.  Although  each  separate  integration  requires  an  arbitrary  con- 
stant, we  write  down  only  a  single  constant  denoting  their  algebraic  sum. 

=  2  a  Cx-^  dx  -  hfx-'dx  +  3  c  fa;*  dx  ^^iJ^T) 

^2a.'^-b.^  +  ^^+C  (by  [4]) 


INTEGRATION  187 


5.  j2ai^-'dx  =  '^^-\-a  6.     C3  7nz'dz  =  ^^+a 


8.  C-V2px  dx  =  I  x^Jpx  +  C, 

9.  C(ai  -x^^dx  =  a^x  - 1-  a^aj^  +  f  a^o;*  -~+0. 
Hint.    First  expand. 

10.  f(a^-f)Wydy=2y^(^^-^-fl+^-^-t^+a 

11.     fC  Va  -  VO'd<  =  ah-2  ati  +  ^  "  ^  +  C' 


12.  f-^,=  inxy  +  c. 

(nx)  " 

13.  f(a'  +  b'l^^xdx  =  ^"-^  +  y)'  +  a. 

Solution.     This  may  be  brought  into  the  form  [4].     For  let 
v  =  a^-\-  bV  and  n  =  ^ ;  then  dv  =  2  Irxdx.     If  we  now  insert 

the  constant  factor  2  6^  before  ccda;,  and  its  reciprocal before 

the  integral  sign  (so  as  not  to  change  the  value  of  the  expres- 
sion), the  expression  may  be  integrated,  using  [4],  namely, 


/ 


,,,71  +  1 

v'^dv  = h  C, 

?i  +  1 


Thus,        j{a'  +  5  V)^-  xdx  =  ^,J{o?  +  h'x'f  2  ft^^xc^a; 
=  2^-,  J(r/  +  h'x'fdia'  +  ?>V) 

262*  I  "^  3^2        -+-    • 

iVb^e.  The  student  is  warned  against  transferring  any  function  of  the 
variable  from  one  side  of  the  integral  sign  to  the  other,  since  that  would 
change  the  value  of  the  integral. 


188  ELEMENTARY   ANALYSIS 

14.  j  Va^  —  x'xdx  =  —  ^(a^  —  o^)^  +  C. 

15.  J (3  ax'  +  4  6.T^)^(2  ax  +  4  /^i^^y^^  ^  1(3  ^^2  ^  4  ^^s^|  _^  ^^ 
//m«.     Use  [4],  making  v  =  S  ax^  -\- 4  bx^  and  w  =  f . 

16.  fb(6  ax'+S  bx')^  (2  ax+A  hx')dx=  — .  (6  ao^^+S  hx^)i  +  C. 
J  16 

Ifiw*.    Write  this    1  (a^  -|-  ys-y^x'^'dz  and  apply  [4]. 

18.  f— ^^=_2vr^^+c 

21-    r|^2  =  V^"  10S(^'  +  '^'«'')  +  ^- 

This  resembles  [5].     For  let  t?  =  6^  +  eV ;  then  dv  =  'l  e'xdx. 
If  we  introduce  the  factor  2  e'  after  the  integral  sign,  and  

before  it,  we  have  not  changed  the  value  of  expression,  and 
the  numerator  is  now  seen  to  be  the  differential  of  the  denomi- 
nator.    Therefore 

=  |^,log(6'^  +  6V)  +  a  (by  [5]) 

22.   J^^=llog(a^2_i)^l^gC'_l(^g^Y^^31. 


INTEGRATION  189 

23.     f^^^P^  =  logc{:^-3a'x)l 
J    OCT —  3  a^x  ^ 

J  8  a  -  6  &«2        s 


(8  a  -  6  6ir)T2 


26.  r^=:a;-|'  +  f-log(a;  +  l)  +  G 
•^   iC  -f- 1  Zi        o 

Hint,     First  divide  the  numerator  by  the  denominator. 

27.  p^^^dx  =  x-\og{2x-3f+a 

28.  f?— =L^  dx  =  -  log  (a;**  -  7ix)  +  C. 
J  X"  —  nx  n 

29.  I  ?^^ ^L-^  =  -\-'±.  -log  9f  ^  a 

J  a  +  hV'      nb     ^^  ^ 

Proofs  of  [6]  and  [7].     These  follow  at  once  from  the  cor- 
responding formulas  for  differentiation,  IX  and  IX  a,  p.  113. 

PROBLEMS 

For  formulas  [6]  and  [7]. 
Verify  the  following  integrations. 


/' 


ba'-dx  ==  -^^  +  C. 


2  log  a 
Solution.      Cba^^dx  =  bCa^^dx.  (by  [2]) 


This  resembles  [6].  Let  v  =  2x',  then  dv  =  2dx.  If  we 
then  insert  the  factor  2  before  dx  and  the  factor  ^  before  the 
integral  sign,  we  have, 

b  Ca'^dx  =  -  Ca'-2  dx  =  -  Ca'^d(2  x)=-  •  -^  +  C. 
J  2 J  2J         ^      ^      2    log  a 

(by  [6]) 


190  ELEMENTARY   ANALYSIS 

3.     C(e''  +  a'^)  dx=^  fe'^  +  -^\  +  G. 

J,    X  X 

e^dx  =  ne""  +  C. 

5.     Ce^"+^^+\x  +  2)dx  =  ^e^"+'^+^  +  a 

n  log  a     m  log  o 

*/  1  +  log  a 

(e«  +  e  ")da;  =  a(e«  —  e  «)  +  (7. 

10.     ("(3  e^*  -  l)h^'dt  =  i  (3  e^*  -  1)*  +  C. 

Proofs  of  [8]-[13].     These  follow  at  once  from  the  corre- 
sponding formulas  for  differentiation,  X,  etc.,  p.  113. 
Proof  of  [_U-\, 

J,         ,    _  rsin^^cg^__       C—  sin  vdv 
J    cos'y  J      cos'y 

/eg  (cos  ^) 
cosv 

=  —  log  COS  v+C  (by  [5]) 

=  log  sec  V  -\-C. 

[Since  —  log  cos  v  =  —  log =  —  log  1  +  log  sec  v  =  log  sec  v. 
sec  V  J 

T>      ^   rr-i'— I       r     4-    ^         Tcos'yd'y       ^(^(sin^?) 
c/  J     sin  V        J     smv 

=  log  sin  v-{-  C.  (by  [5]  ) 


INTEGRATION  191 

PROBLEMS 

For  formulas  [8]-[17]. 

Verify  the  following  integrations. 

1.     I  sm  2  axdx  = [-  (7. 

c/  2a 

Solution.      This    resembles    [8].      For    let   v  =  2  ax,    then 
dv  =  2  adx.     If  we  now  insert  the  factor  2  a  before  dx,  and 

the  factor  —  before  the  integral  sign,  we  get 
2a 


J  sin  2  axdx  =  —  i  sin  2  ax -2  adx 
2aJ 


1  r  1 

=  — -  I  sin  2  ax  -  d  (2  ax)  =  - —  —  cos  2ax-\-  C 
^''^  ^"^  (by  [8]) 

cos  2ax      p 


2a 


2.     I  cos  mxdx  =  —  sin  mx  +  C. 
J  m 

5  sec^  bxdx  =  -  tan  bx  +  C 

4.  r/'cos  ^  -  sin  3  (9"]  d0  =  3  sin  ^  +  ^  cos  3  (9  +  (7. 

5.  I  7sec3«tan3  acZa  =  |sec3a4- C 

6.  I  A:  cos  (a  +  57/)  dy  =  -  sin  (a  +  by)  +  C. 

7.  I  cosec^  x^  •  cc^dT  =  —  1  cot  a^  +  (7. 

8.  I  4  CSC  ax  cot  aoJcfiK  = esc  ax  +  (7. 

*/  a 

^      C   sin  a*da?         ^ 

9.  I =  log 


a  -f  &  cos  aj  (a  -f  6  cos  a;)^ 


192  ELEMENTARY  ANALYSIS 

10.     I  e'^'''  sin  xdx  =  —  e''^'^  +  C. 

11-    r      2/''    ,      =-J:taii(a-&a^)+a 
*/  cos^  (a  —  6^:1:^)  6 

cos  (log  x)  —  =  sin  log  x  +  (7. 

X 

13.  f^^  =  _ncot^  +  a 
^  sin^^*  ^ 

14.  I  ^^ — ^ ^ —  =  log (x  +  sm  a;)(7. 

c/         a;  +  sm  0? 

15.  r?iB4^  =  sec<^  +  a 

J      COS-  </) 

16.  I  sm^xdx  =  ^x  —  ^sm2x+C, 

[Substitute  sin^ic  =  1  —  1  cos  2  cc.] 

17.  j  cos^ a^c?i»  =  ^ X -\- 1 sm2 X -i-  O, 
[Substitute  cos^  x  =  J  +  J  cos  2  a:.] 

18.  I  tan^  xdx  =  tan  x  —  x-\-0, 
[Substitute  tan'^  x  =  sec^  x  —  1.] 

19.  \  Qot!-^  xdx  =  —  cot  X  —  X  +  C. 

[Substitute  cot^  x  =  cosec^  x  —  1.] 
Proof  of  [^16^.     Since 

d  fare  tan  -  +  C )  =  — ^^  =  ^^„  (by  XVIT,  p.  114) 


1+     - 

we  get  I    /'^  ^  =  -  arc  tan  -  +  (7. 

*/  ^^''  +  a-      a  a 


INTEGRATION  193 

Proof  of  117].     Since 


d 


d  I  arc  sin  --{-C]=     ,    ^^^      =  --^=,  (by  XVI,  p.  114) 


we  get       I  — ^zz^zz  =  arc  sm  -  +  C . 


PROBLEMS 


For  formulas  (16)  and  (IT). 
Verify  the  following  integrations. 

C    dx  1        ,      2x  ,   ^ 

1.     I =  -  arc.  tan \-  C. 

J  Ax'-}-9      6  3 

Solution.  This  resembles  [16].  For,  let  v^  =  4:  x^  and  a-  =  9 ; 
then  V  =  2  x,  dv  =  2  dx,  and  a  =  3.  Hence  if  we  multiply  the 
numerator  by  2,  we  get 

r     dx      ^  1  r       2dx       ^  1  r     d(2x) 

J  4.x' +  9      2 J  (2  xf  +  (3)2     2 J  (2  xf  +  (^f 

=  Jarctan|?+a  (by  (16)) 

0  3 


2.    I  —  -  arc  sm \- C, 

J  V1  (>  -  9  X?      3  4 


_.       i      o  xctx  O  •        0   1    /^ 

3.    I  — =:  =  -  arc  sm  x-  +  (7. 
J  VI  -  «*     2 

4-    r^  =  ^arctan^  +  C 

JV9-4a;2     2  3 

6.  f—i^^arcsin-^+C. 

„      r      7  ds  7  .       /5      ,   ^ 

7.  I  —  = arc  sin-\/~s  + C. 


COS  ada        1 

=  -  arc 


a^  +  siii^  a      a 
e*dt 


t^f'^^a 


Vl  -  e'' 
dx 


\    a 
arc  sin  e*  +  (7. 


194  ELEMENTARY   ANALYSIS 

10.     I "^"^  =  arc  sin  (log  a^)  +  O. 

^u  +  &^ 


11. 


a;  Vl  —  log^  X 

/du 
Va2-(^  +  6)^ 

12.     f -^^^ 


:  arc  sm 


+  a 


arc  tan  — ^ h  (7. 


74.  Trigonometric  and  other  substitutions.  A  useful  de- 
vice for  integration  is  afforded  by  simple  transformation  of  the 
variable.     The  following  examples  illustrate  this  statement. 


EXAMPLES 


1.    Work  out 


/: 


X  (XX 


Hence    f^^^^  f 


Va^  —  01? 
Solution.     Substitute 
x=^a  sin  0. 
,'.  dx=a  cos  0  dOj 

V«^  —  x^  =  -y/ct^  —  a^  sin^  0  =  a  cos  0. 
a^  sin^  0  •  a  cos  0  •  dO 


a  cos  0 
a2[|(9-i^sin2(9]. 


=  a^  I  SI 


sin2  (9  dO 


From  the  figure,  ^  =  arc  sin  -, 

a 

sin  2  ^  =  2  sin  0  cos  0  =  2 


(problem  16,  p.  192) 

x    -y/a?  —  a? 


r     a?dx      ^ 

^    Vc6^  —  X^ 


—  arc  sin -x  Va^  —  x^  +  G.     Ans, 

2  a     2 


Integrals  inyolving  the  radical  Va'^  —  oc'^  may  be  worked  in  this 
way. 


INTEGRATION  195 

2.    Work  out 


Solution.     Substitute  aj  =  -. 

y 


\  da;  = ^.     Also  Va^-f  ^  =\/«  +-^  = '^  ° 

2/'  ^       f  y        ' 

J_dy 
y'      ^_  r     ?/# 
l.VaV  +  1         -/  VaV  +  1* 


Hence  

T  y 


=  -  V  ^''V'  +  ^  (by  (4)) 


a^ 


>^^l+^+a    ^n.. 


-PROBLEMS 
Work  out : 


1.        f ^^     ^_V^'-^  +  a  rputa.=i.' 

2.  fVo^^^  dx  =  ^  Va^^^  +  ^  a'  arc  sin  -  +  (7. 
c/  2  2  a 

3.  r       ^^        =  - 1  arc  sin « +  C.  [Put  a;  =  ^ 
5.               r-^^^  =  -^-arcsin^  +  G 


6.  rVZEZ^^^_VZEZ_arcsm^  +  a     • 

J        x^  X  a 


CHAPTER   XIV 

CONSTANT   OF  INTEGRATION 

75.  Determination  of  the  constant  of  integration  by  means 
of  initial  conditions.  In  order  to  determine  the  constant  of 
integration,  data  must  be  given  in  addition  to  the  differential 
expression  to  be  integrated.  Let  us  illustrate  by  means  of 
examples. 

EXAMPLES 

1.  Find  a  function,  whose  first  derivative  is  3x'^  —  2x  +  5y 
and  which  shall  have  the  value  12  when  x  =  l. 

Solution.  (Sx^~-2x-^o)dx  is  the  differential  expression  to 
be  integrated.     Thus, 

C(3x^-2x  +  5)dx  =  x^-x^  +  5x+C, 

where  C  is  the  constant  of  integration.     From  the  conditions 
of  our  problem  this  result  must  equal  12  when  x  =  l]  that  is, 

12  =  l_l+5  +  C,  or  0=7. 

Hence  x"  —  x'^  --  5x  -\-7  is  the  required  function. 

2.  Find  the  laws  governing  the  motion  of  a  point  which 
moves  in  a  straight  line  with  constant  acceleration. 


Solution.     Since  the  acceleration 
is  constant,  say/,  we  have 


^^^  from  (6),  Art.  6T~] 


dt 


'dt~-'' 

or 

dv  —fdt.     Integrating, 

(-1) 

v=ft+C. 

196 


CONSTANT  OF  INTEGRATION  197 

To  determine  C,  suppose  that  the  initial  velocity  be  Vq  ;  that 
^^^  ^^^  v=zVq  when  ^  =  0. 

These  values  substituted  in  {A)  give 
^y^  =  0  +  (7,  or  C=  Vq. 

Hence  {A)  becomes 
(B)  v=ft  +  v, 


'0- 

ds 
Since  v  =  —  [Art.  67],  we  get  from  (B) 

(Xv 

ds     J,.  , 
dt 

or  ds=ftdt-\-Vodt,     Integrating, 

(G)  s  =  ^ft'-\-Vot+a 

To  determine  (7,  suppose  that  the  initial  space  (=  distance) 
be  Sq  ;  that  is,  let 

s  =  Sq  when  ^  =  0. 

These  values  substituted  in  (C)  give 

S(3  =  0  +  0  +  C,  or  (7  =  So. 
Hence  (C)  becomes 
(D)  s  =  ^ft'-\-v,t  +  So. 

Formulas  (B)  and  (D)  are  the  required  laws. 

3.    A  certain  magnitude  z  varies  with  the  time  according  to 

dz 
the  Compound  Interest  Law,  that  is,  z  and  — ^  are  proportional. 

Find  z  as  a  function  of  the  time  t. 
Solution.     By  definition, 

dz 

—  =  az, 
dt         ' 

where  a  is  a  constant  factor  of  proportionality.     Multiplying 
both  sides  by  dt  and  dividing  through  by  z  gives 

—  =  adt. 

z 


198  ELEMENTARY   ANALYSIS 

Integrating, 

(1)  log  z  =  at^C. 

To  determine  C,  assume  that  the  value  of  z  when  t  =  0  \i 
denoted  by  %     Substituting  in  (1),  C=\ogZo. 
We  may  now  write  (1)  in  the  form 

(2)  log  z  —  log  Zq  =  at,  or  log  —  =  at. 
Hence,  changing  to  exponentials  (Art.  33), 

—  =  e«^^  or  ;^  =  ZQe""^..    Ans, 

PROBLEMS 

Find  the  function  whose  first  derivative  is 

1.  X  —  3,  knowing  that  the  function  equals  9  when  x  =  2. 

Jns.  ?^_3a;  +  13. 
2 

2.  3  -\-x—  5 x^,  kuQwing  that  the  function  equals  —  20  whei 
^"=^-  Ans.  124  +  3^;+--  — 

3.  (y^  —  b^y)dy,  knowing  that   the   function   equals  0  whei 

y  =  ^-  Ans.  y--^^  +  2b^-4. 

4        2 

4.  sin  a  +  cos  cc,  knowing  that  the  function  equals  2  whei 
cc  =  -'  Ans.  sin  a  —  cos  a  + 1. 

Find  the  equation  of  a  curve  such  that  the  slope  of  th( 
tangent  at  any  point  is 

5.  Sx-2.  Ans.  y  =  '^-2x+a 

6.  ':)i?-\-^x,  the  curve  passing  through  the  point  (0,  3). 

Ans.  2/  =  -  +  — +  3. 


CONSTANT   OF   INTEGRATION  199 

7.  — ,  the  curve  passing  through  the  point  (0,  0). 

y 

Ans.  y^  =  2px, 

8.  — ,  the  curve  passing  through  the  point  (a,  0). 

ay 

Ans.  bV  —  a^y^  =  a^b^. 

9.  m,  the  curve  making  an  intercept  b  on  the  axis  of  y, 

Ans,  y  =  mx  +  b. 
Assuming  that  v  =  Vo  when  ^  =  0,  find  tlie  relation  between  v 
and  t,  knowing  that  the  acceleration  is 

10.  Zero.  Ans,  v  =  Vo. 

11.  Constant  =  A;.  Ans.  v=Vi)  +  lct. 

bf 

12.  a  +  bt.  A71S.  v  =  Vo  +  at-] 

Assuming  that  s  =  0  when  ^  =  0,  find  the  relation  between  s 
and  t^  knowing  that  the  velocity  is 

13.  Constant  (=  t'o)-  Ans.  s  =  Vot. 

nf 

14.  m  +  nt.  Ans.  s  =  mt-\ 

.15.   3  +  2^-3^1  Ans.  s  =  3t  +  f-f. 

16.  The  velocity  of  a  body  starting  from  rest  is  5f  ft.  per 
second  after  t  sec.  (a)  How  far  will  it  be  from  the  point  of 
starting  in  3  sec.  ?  (b)  In  what  time  will  it  pass  over  a 
distance  of  360  ft.  measured  from  the  starting  point  ? 

A71S.   (a)  45  ft. ;  (b)  6  sec. 

17.  A  train  starting  from  a  station  has  after  t  hr.  a  speed 
of  f  —  21  ^^  +  80  ^  mi.  per  hour.  Find  (a)  its  distance  from 
the  station;  (b)  during  what  interval  the  train  was  moving 
backwards ;  (c)  when  the  train  repassed  the  station ;  (d)  the 
distance  the  train  had  traveled  when  it  passed  the  station  the 
last  time.  Ans.   (a)  ^t^ —  T  f -\-4.0f  mi. ; 

(b)  from  5th  to  16th  hr. ; 

(c)  in  8  and  20  hr. ; 

(d)  46581  mi. 


200  ELEMENTARY   ANALYSIS 

18.  The  equation  giving  the  strength  of  the  current  i  for  the 
time  t  after  the  source  of  E.M.F.  is  removed,  is  {R  and  L  being 
constants) 

dt 

_m 

Find  I,  assuming  that  J=  current  when  ^  =  0.      Arts,  i=:  le  ^. 

19.  Find  the  current  of  discharge  i  from  a  condenser  of 
capacity  (7  in  a  circuit  of  resistance  R,  assuming  the  initial 
current  to  be  Iq,  having  given  the  relation 

di  _  dt 

J'^CR' 

t 

C  and  R  being  constants.  Ans.  i  =  I^e^^, 


M  N 


CHAPTER   XV 

THE  DEFINITE  INTEGRAL 

76.   Differential  of  an  area.     Consider  the  curve  AB  whose 
equation  is 

Let  CD  be  a  fixed  and  MP  a  vari- 
able ordinate,  and  let  u  be  the 
measure  of  the  area  CMPD.* 
When  X  takes  on  a  sufficiently 
small  increment  Aa;,  u  takes  on 
an  increment  A?^  (=  area  JOTQP). 
Completing  the  rectangles  MNRP  and  MNQS,  we  see  that 

areaJf^i^P<  area  MNQPk  area  MNQS, 

oi:  JfP.  Aa;<Aifc<^Q.  Ao;; 

dividing  by  Ao;, 

MP<^<Nq.\ 

Aa; 

Now  let  ^x  approach  zero  as  a  limit  j  then  since  MP  remains 
fixed  and  NQ  approaches  MP  as  a  limit,  we  get 


or,  using  differentials. 


du  =  2/^^' 


*  We  may  suppose  this  area  to  be  generated  by  a  variable  ordinate  start- 
ing out  from  CD  and  moving  to  the  right  ;  hence  u  will  be  a  function  of  ic, 
which  vanishes  when  x  —  a. 

t  In  this  figure  MP  is  less  than  NQ  ;  if  MP  happens  to  be  greater  than  NQ, 
simply  reverse  the  inequality  signs. 

201 


202 


ELEMENTARY   ANALYSIS 


Theorem.  The  differential  of  the  area  hounded  by  ayiy  curve, 
the  axis  of  X,  and  a  fixed  and  a  variable  ordinate  is  equal  to  the 
product  of  the  variable  ordinate  and  the  differential  of  the  corre- 
sponding abscissa, 

77.  The  definite  integral.  It  follows  from  the  theorem  in 
the  last  section  that  if  AB  is  the  locus  of 


then 


du  =  ydXy  or 
du  =  ^{x)dx 


Y 

^^ 

\ 

1 

/^ 

m 

\ 

2jm 

f 

^8 

■1 

^p 

0 

iraJ(_ 

M. 
-b-— 

E    X 

>j 

is  the  differential  of  the  area  between  the  curve,  the  axis  of  x, 
and  two  ordinates.     Integrating  {A), 

u  =  I  <^{x)dx. 

Let  j  ^{x)dx  be  worked  out  and  denote  the  result  hjfix)  +  C, 

(B)  r.u=f{x)  +  a 

We  may  determine  (7,  as  in  Chapter  XIII,  if  we  know  the 
value  of  ^i  for  some  value  of  x.  If  we  agree  to  reckon  the  area 
from  the  axis  of  y,  i.e.  when 

(C)  x  =  a,u  =  2iVQ^  OCDG, 

and  when  x=zb,  u  =  area  OEFG,  etc., 

it  follows  that  if 

(D)  x=0,  then  i6  =  0. 
Substituting  (Z>)  in  (5), 

0=/(0)  +  C,orC=-/(0). 


THE   DEFINITE   INTEGRAL  203 

Hence  from  (B)  

(E)  u  =f(x)  -/(O), 

giving  the  area  from  the  axis  of  y  to  any  ordinate  (as  MP), 

To   find   the   area  between  the  ordinates  CB  and   EF^  we 
observe  that 

(F)  area  GEFD  =  area  OEFG  -  area  OCDG, 
But,  using  (E), 

area  OCDG  =f(a)  -/(O), 
area  OEFG  =f(b)  -/(O). 
Substituting  in  (F), 

(G)  area  CEm  =  f(fi)  -  f{a). 

Theorem.      Tlie  difference  of  the  values  of  I  ydx  for  x  =  a  and 

x  =  b  gives  the  area  hounded  by  the  curve  whose  ordinate  is  y,  the 
axis  of  Xj  and  the  ordinates  corresponding  to  x  =  a  and  x  =  b. 
This  difference  is  represented  by  the  symbol 

ydxj  or    i    cl)(x)dx, 

and  is  read  ^^  the  integral  from  a  to  &  of  ydx.^^  The  operation 
is  called  integration  between  limits,  a  being  the  lower  and  b  the 
upper  hmit. 

Since  {H)  always  has  a  definite  value,  it  is  called  a  definite 
integral.     For,  if 

I  cf){x)dx  =  f(x)  +  C, 


f{h)+C 


then     Cci,{x)dx  =  UXx)  +  O 

or  j    cl,(x)dx=f(b)-f(a), 

the  constant  of  integration  having  disappeared. 


/(a)  +  C 


204  i:lementary  analysis 

We  may  accordingly  define  the  symbol 

<p(x)dx 


as  the  numerical  measure  of  the  area  bounded  by  the  curve  y  =  <l>(x), 
the  axis  of  X,  and  the  ordinates  of  the  curve  at  x=a,  x=b.  This 
definition  presupposes  that  these  lines  bound  an  area,  i.e.  the  curve 
does  not  rise  or  fall  to  infinity,  and  both  a  and  b  are  finite, 

78.  Geometrical  representation  of  an  integral.  In  the  last 
section  we  represented  the  definite  integral  as  an  area.  This 
does  not  necessarily  mean  that  every  integral  is  an  area,  for  the 
physical  interpretation  of  the  result  depends  on  the  nature  of 
the  quantities  (j>(x)  and  x.  In  the  chapter  on  functions,  the 
variable  x  .was  chosen  to  represent  magnitudes  of  various 
kinds  —  time,  length,  etc.  The  corresponding  function  might 
be  a  volume,  or  any  other  physical  magnitude.  The  definite 
integral  {H)  is  then  represented  in  numerical  value  by  the 
area  in  the  graph  of  <^(a;),  but  its  actual  physical  significance 
might  be  something  quite  different.  For  example,  if  we  turn 
to  equation  (6),  p.  168,  and  represent  the  acceleration  as  a 
function  of  the  time  by  a  {t),  then,  multiplying  through  by  dt 

and  integrating,  we  obtain  the  indefinite  integral  v  =  |  a(t)dt. 

The  difference  of  two  values  of  this  integral,  that  is,  the  defi- 
nite integral,  is  clearly  a  change  in  velocity,  and  hence  is  a 
velocity.     Further  illustrations  occur  in  later  sections. 

79.  Calculation  of  a  definite  integral.  The  process  may 
be  summarized  as  follows: 

First  step.  Find  the  indefinite  integral  of  the  given  differential 
expression. 

Second  step.  Substitute  in  this  indefinite  integral  first  the 
tipper  limit  and  then  the  lower  limit  for  the  variable,  and  subtract 
the  last  result  from  the  first. 

It  is  not  necessary  to  bring  in  the  constant  of  integration,^ 
since  it  always  disappears  in  subtracting. 


THE   DEFINITE   INTEGKAL 
EXAMPLES 


205 


1.  Find   r  x^dx, 

x^dx=   ^ 

1  [_o 

2.  Find    |     ^inxdx. 
Solution.      I    sin  xdx  = 

3.  Find   r^^,- 
Solution.       I    -— : 


4      64      1 


:  21.        ^TIS. 


—  cos  a; 


Jo 


"1  .  x' 
-  arc  tan  - 
a  a 


Ans. 


1  ± 

-  —  arc  tan  1 arc  tan  0 

a  a 

3^_0  =  ^.     Ans. 
4a  4a 


PROBLEMS 
3. 


1.  r6x'dx  =  3S.  3.    P— =  1. 

2.  I    (a-o;  —  a;^^?^  =  T-  •  4.1    —  =1. 

Jo    ^  '4  Ji     X 

6.     f        ^-^        =V3-1. 

Jo  V3-2a; 


n_tdt_ 

"Jo      aH 


lo,£r2 


'^^       da; 


V2-3a;=^      4V3 


9. 


10. 


'^     3  xdx 


</l25. 


Jo  2/2  -  2/ +  1      3V3 


+  ^^         2 

TT 

13.     I     sin<^cZ<^  =  l. 

X  +  QO 
e~=^  cZo;  =  1. 

J2r  V2r" 
Vo; 


206  ELEMENTAKY  ANALYSIS 

Jo  Vr-  — a;-       ^  *^o     V2r  — ?/ 

J_ja*^^         ^     ^       315  a< 
20.  2a  P(2+2cos^)*di9=8ci.   21.    |    t^nada  =  0. 


CHAPTER   XVI 

INTEGRATION  A  PROCESS  OF   SUMMATION 

80.  Introduction.  Thus  far  we  have  defined  integration  as 
the  inverse  of  differentiation.  In  a  great  many  of  the  applica- 
tions of  the  Integral  Calculus,  however,  it  is  preferable  to 
define  integration  as  a  process  of  summation.  In  fact  the  In- 
tegral Calculus  was  invented  in  the  attempt  to  calculate  the 
area  bounded  by  curves  by  supposing  the  given  area  to  be 
divided  up  into  an  "  infinite  number  of  infinitesimal  parts 
called  elements,  the  sum  of  all  these  elements  being  the  area 
required."  Historical!}^  the  integral  sign  is  merely  the  long 
S,  used  by  early  writers  to  indicate  "  sum.'^ 

This  new  definition,  as  amplified  in  the  next  section,  is  of 
fundamental  importance,  and  it  is  essential  that  the  student 
should  thoroughly  understand  what  is  meant  in  order  to  be 
able  to  apply  the  Integral  Calculus  to  practical  problems. 

81 .  The  fundamental  theorem  of  the  Integral  Calculus.     If 

<i>{x)  is  the  derivative  oi  f{x),  then  it  has  been  shown  in  §  77, 
p.  203,  that  the  value  of  the  definite  integral 

{A)  (\(p)dx==f{h)-f{a) 

gives  the  area  bounded  by  the  curve  y  =  <^(a^),  the  o^axis,  and 
ordinates  erected  at  a;  =  a  and  x—h. 

Let  us  now  make  the  following  construction  in  connection 
with  this  area. 

Divide  the  interval  from  x  =  atox  =  h  into  any  number  n  of 
equal  subintervals,  erect  ordinates  at  the  points  of  division,  and 
complete  rectangles  by  drawing  horizontal  lines  through  the 

'207 


208 


ELEMENTARY   ANALYSIS 


6 ^ 


extremities  of  the  ordinates,  as  in  the  figure.     It  is  clear  that 

the  sum  of  the  areas  of 
^  these    n    rectangles   (the 

shaded  area)  is  an  ap- 
proximate value  for  the 
area  in  question.  It  is 
further  evident  that  the 
limit  of  the  sum  of  the 
areas  of  these  rectangles, 
when  their  number  n  is 
indefinitely  increased,  will  equal  the  area  under  the  curve. 

Let  us  now  carry  through  the  following  more  general  con- 
struction. Divide  the  in- 
terval into  n  subintervals, 
not  necessarily  equal,  and 
erect  ordinates  at  the 
points  of  division.  Choose 
a  point  within  each  sub- 
interval  in  any  manner, 
erect  ordinates  at  these 
points,  and  through  their 
extremities  draw  horizon- 
tal lines  to  form  rectangles  as  in  the  figure.  Then,  as  before, 
the  sum  of  the  areas  of  these  n  rectangles  equals  approximately 
the  area  under  the  curve,  and  the  limit  of  this  sum  as  n  in- 
creases without  limit  and  each  subinterval  approaches  zero  as 
a  limit  is  precisely  the  area  under  the  curve. 

These  considerations  show  that  the  definite  integral  {A)  may 
be  regarded  as  the  limit  of  a  sum.  Let  us  now  formulate  this 
result. 

1.  Denote  the  lengths  of  the  successive  subintervals  by 

Aa^i,  Aa?2,  Ai»3,  ...,  ^x^. 

2.  Denote  the  abscissas  of  the  points  chosen  in  the  subinter- 
vals by 

Xi,  X2,  Xq,  •••,  x^. 


INTEGRATION   A   PROCESS   OF   SUMMATION       209 


Then  the  ordinates  of  the  curve  at  these  points  are 

</>(^l),    ^(^2),    <^(^3),    •••,    <l>(^n)' 

3.  The  areas  of  the  successive  rectangles  are  obviously 

4.  The  area  under  the  curve  is  therefore  equal  to 


limit 
n=co 


(l>(x^)Ax^+  <I>(X2)AX2+  •••  + 
The  discussion  gives  the  equation 
(B)  j\{x)dx  =  ]^l  \<i>{x,)Ax,  +  <t>{x,)Ax, 


ct>{x,,)Ax^ 


+ 


+  <t>{x„)Ax^ 


The  equation  (B)  has  been  established  by  making  use  of  the 
notion  of  area.  The  area  under  discussion  is  bounded^  that  is, 
it  has  a  closed  perimeter  consisting  of  the  curve  y  =  <^(ic),  the 
ic-axis,  and  the  lines  x  =  a,  x  =  b.  It  is  therefore  tacitly 
assumed  that  the  function  (l>(x)  is  finite  for  all  values  of  a;  from 
a  to  6  inclusive.  That  is,  the  curve  y  =  <i>  {x)  does  not  run  off 
to  infinity  between  ic=«  and  x^h]  otherwise  expressed,  the 


210  ELEMENTARY  ANALYSIS 

curve  has  no  "  breaks  "  in  it.  The  student  may  refer  to  Art. 
32  for  examples  of  curves  which  have  *^  breaks ''  in  them  ;  that 
is,  have  vertical  asymptotes.  The  property  of  the  function 
</) (x)  here  described,  that  is,  of  remaining  finite  between  x=  a 
and  x  =  h,  and  yielding  a  continuous  graph,  is  expressed  by  the 
word  "  continuous.^' 

Intuition  has  aided  us  in  establishing  the  result  (B).  Let 
us  now  regard  (B)  simply  as  a  theorem  in  analysis,  which  may 
be  stated  thus : 

Fundamental  theorem  of  the  Integral  Calculus.  Let  cl>(x)  be 
continuous  for  the  interval  x  =  a  to  x=:b.  Let  this  interval  he 
divided  into  n  subintervals  whose  lengths  are  AXi,  Aa?27  '"y  ^^n>  ^^^ 
let  points  be  chosen,  one  in  each  subinterval,  their  abscissas  being 
Xi,  a?2,  •••,  x^  respectively.     Consider  the  sum 

(O)  (^ (x{)  Ao^i ^cl>(x2)Ax2  +  "'  +<l> (x^) Ax^. 

Then  the  limiting  value  of  this  sum  ivhen  n  increases  without 
limit  and  each  subinterval  approaches  zero  as  a  limit  equals  the 

value  of  the  definite  integral  i    <^  (x)  dx. 

The  importance  in  the  application  of  this  theorem  is  this : 
We  are  able  to  calculate  by  integration  any  magnitude  which  is 
the  limit  of  a  sum  of  the  form  (C). 

The  difficulty  which  arises  in  actual  practice  is  that  of  de- 
termining the  function  <^(a?).  No  general  rule  applicable  to 
all  cases  can  be  given  for  overcoming  this  difficulty.  Th^ 
student  should  study  the  examples  worked  out  in  the  follow- 
ing pages  in  order  to  obtain  practice. 

We  may  observe  that  the  terms  of  the  sum  (C)  are  of  the 
form 

{B)  cjy  (x)  Ax,  or  also  <^  (x)  dx  (since  dx  —  Ax), 

which  may  be  called  the  general  term,  or  also  an  element  of  the 
integral. 

The  following  directions  will  be  found  useful  in  applying 
the  Fundamental  Theorem. 


INTEGRATION   A   PROCESS   OF   SUMMATION       211 


First  step.  Divide  the  required  magnitude  into  similar  parts 
such  that  it  is  clear  that  the  desired  result  will  be  found  by 
summing  up  these  parts  and  passing  to  the  limit. 

Second  step.  Choose  a  suitable  variable  such  that  the  mag- 
nitude of  each  part  can  be  expressed  in  the  form  (D). 

Third  stejj.     Apply  the  Fundamental  Theorem  and  integrate. 

82.  Areas  of  plane  curves.  As  already  explained,  the  area 
between  a  curve,  the  axis  of  X,  and  the  ordinates  x  =  a  and 
x  =  b  is  given  by  the  formula 


(^) 


area 


j^  y  doc, 


the  value  of  y  in  terms  of  x,  being  substituted  from  the 
tion  of  the  curve. 

Equation  {A)  is  readily  memo- 
rized by  observing  that  the  inte- 
grand ydx  represents  the  area  of  a 
rectangle  CR  of  base  dx  and  alti- 
tude y. 

The  a^Dplication  of  the  Funda- 
mental Theorem  to  the  calculation 
of  the  area  bounded  by  the  curve 


equa- 


(1) 


x  =  4>{y\ 


the  2/-axis  and  abscissas  at  2/  =  c  and  y  =  d,  is  immediate. 

First  step.     Construct  the  n  rectangles 
as  in  the  figure.     The  area  is  clearly  the 
limit  of  the  sum  of  these  rectangles  as 
their  number  increases  indefinitely. 
Second  step.     Call  any  one  of  the  alti- 
^x=4>{y)   tudes  A?/.     The  base  of  any  one  of  the 
rectangles  is  x.     Hence  the   general  ex- 
pression for  the  area  of  the  rectangles  is 
X  ^y   or  </)(?/)  A^V?   from  (1).     This   is   an 
element  of  the  required  integral. 


212 


ELEMENTARY   ANALYSIS 


Third  step.     Applying  the  Fundamental  Theorem  gives  the 
formula 


area: 


''So  "^^y^^y- 


The  result  may  be  stated :    The  area  between  a  curve,  the 
y-axis,  and  abscissas  2X  y  =  G  and  y  =  di^  given  by 


iB) 


area 


the  value  of  x  in  terms  of  y  being  found  from  the  equation  of 
the  curve. 

EXAMPLES 

1.  Find  the  area  included  between  the  semicubical  parabola 
?/2  =  aj3  ^^^  |.]^g  2 jj^g  a;  =  4. 

Solution.  Let  us  first  find  the  area  OMP,  half  of  the  re- 
quired area  OPP'.  For  the  upper  branch  of  the  curve  y  =  -Va^, 
and  summing  up  the  rectangles  between  the  limits  x  =  0  and 
X  =  4,  we  get,  by  substituting  in  (A), 


area 


OMP  =  Cydx  =  Cxhx  =  -6-^4  =  124. 


-V-  =  25f. 


Hence  area  OPP'  =  2 

If  the  unit  of  length  is  one  inch,  the  area 
of  OPP  is  25fsq.  in. 

Note.    For  the  lower  branch  y=—  x^ ;  hence 

area  OMP  =  (\~x^)dx  =  -\2  -  f . 

Tills  area  lies  below  the  axis  of  x  and  has  a 
negative  sign  because  the  ordinates  are  negative. 

In  finding  the  area  OMP  above,  the  result  was  positive  because 
the  ordinates  were  positive,  the  area  lying  above  the  axis  of  x. 

The  above  result,  25f,  was  the  total  area  regardless  of  sign. 
As  we  shall  illustrate  in  the  next  example,  it  is  important  to 
note  the  sign  of  the  area  when  the  curve  crosses  the  axis  of  X 
within  the  limits  of  integration. 


INTEGRATION  A  PROCESS  OF   SUMMATION      213 

2.  Find  the  area  of  one  arch  of 
the  sine  curve  y  =  sin  x. 

Solution.       Placing    ^  =  0    and 
solving  for  x,  we  find  -^ 

ic  =  0,  TT,  2  TT,  etc. 

Substituting  in  {A),  p.  211, 

area  OAB  =  I    ydx  =  I      sin  xdx  =  2. 

Jr»6  /»27r 

I    ydx  =  I      sin  xdx  =  —  2, 

and        area  OABCD  =  I    ydx=  i      sin  xdx  =  0. 

This  last  result  takes  into  account  the  signs  of  the  two  sepa- 
rate areas  composing  the  whole.  The  total  area  regardless  of 
these  signs  equals  4. 

3.  Find  the  area  included  between  the  parabola  x^  =  4:ay 
and  the  witch 


y- 


xF  +  4za^ 


Solution.  To  determine  the  limits  of  integration,  we  solve 
the  equations  simultaneously  to  find  where  the  curves  intersect. 
The  coordinates  of  A  are  found  to  be  (—2  a,  a),  and  of  B 
(2  a,  a). 

It  is  seen  from  the  figure  (p.  214)  that 

•  area  AOCB  ==  area  DECBA  -  area  DECOA, 

But  area  DECBA  =  2  x  area  0ECB  =  2  C^  ^^'f^,  =  2  vra^, 

Jo     aj2  4-  4  a^ 

and      area  DECOA  =  2  x  area  DEC  =2  r^^dx  =  ^^^ 

Jo     4  a 


Hence area^OO^  =  2  ira^- ^=2  a' fir-  - ] •     Arts. 


3 

\ .       An 

3> 


214 


ELEMENTARY   ANALYSIS 


Another  method  is  to  consider  the  strip  PS  as  an  element  of 
the  area.  If  y'  is  the  ordinate  corresponding  to  the  witch,  and 
y"  to  the  parabola,  the  differential  expression  for  the  area  of 
the  strip  FS  equals  (y'  —  y")dx.  Substituting  the  values  of 
y'  and  y^'  in  terms  of  x  from  the  given  equations,  we  get 

area  AOOB  =  2  x  area  OCB 

■•2a 


-4 


'0     V  o;^  +  4  a^ 


x'' 


4taJ 


\dx 


:2a2(7r-|). 

X        u 
"4.   Find  the  area  of  the  ellipse  —  +  ^  =  1. 

Solution.     To  find  the  area  of 

the  quadrant  OAB,  the  limits 

are  a?  =  0,  ic  =  a ;  and 

h, 

a 

Hence,  substituting  in  (^), 

p.  211, 

area  OAB  =  ~  f^w'-a^^Ux 
aJo 

VhX  ^    o  2\l    ,    ct^ 


y- 


0      /~2 


I  bx  .  2 


arc  sm- 

2  a 


nrdb 


(problem  2,  §  74) 


Therefore  the  entire  area  of  the  ellipse  equals  it  ah. 
We  may  also  calculate  the  integral  giving  area  Q^^as  follows. 
Let  a;=acos  <^  (see  figure  and  §  74).    Then  -y/a^—x^^a  sin  ^, 
dx=z  —  a  sin  <^ d<j>. 

Changing  the  limits  :  —  when  a;=0,  <^=  |^7r;  when  aj=a,  <j5)=0. 


:.  ^  C(a^  -  x'^)^dx  =-abC  sin^  ^  d<^ 

aJo  J^^n 

,r^_lsin2<^ 
[2      4  _ 


=z  —ab\ 


=  '^^^^   as  before. 
4 


INTEGKATION  A   PROCESS  OF   SUMMATION      215 


PROBLEMS 

1.  Find  the  area  bounded  by  the  line  y  =  Bx^  the  a;-axis, 
and  x==:2.  Ans.  10. 

2.  Find  the  area  bounded  by  the  parabola  ?/  =  4:X,  the  axis 
of  X,  and  the  lines  cc  =  4  and  x=9.  Ans.  25^. 

3.  Find  the  area  bounded  by  the  parabola  y^  =  4zX,  the  axis 
of  Y,  and  the  lines  2/  =  4  and  y  =  6.  Ans.  12|. 

4.  Find  the  area  of  the  circle  x"^  -\-y^  =  ?^.  Ans.  irr^, 

5.  Find  the  area  between  the  equilateral  hyperbola  xy  —  a-, 
the  axis  of  X,  and  the  ordinates  x  =  a,  x  =  2  a.      A7is.  a^  log  2. 

6.  Find  the  area  between  the  curve  ^  =  4  —x^  and  the  axis 
of  X.  Ans.  lOf. 

7.  Find  the  area  intercepted  between  the  coordinate  axes 
and  the  parabola  x^  -\-y^  =  a^.  Ans.  i  a^. 

8.  Find   the   area    bounded  by  the   semicubical  parabola 
y^  =  x%  the  axis  of  Y,  and  the  line  2/ =  4.  Ans.  f^l024. 

9.  Find  the  area  between  the  catenary  y  z=  --r^a  _^  ^   ai  ^j^g 
axis  of  Y,  and  the  line  x  =  a.  Ans.   —  [e^  —  1]. 

10.  Find  the  area  between  the  witch  y  = and  the 

^       x'  +  ^a^ 

axis  of  Xy  its  asymptote.  Ans.  4  ira^. 

11.  Find  the   area   included    between    the   two   parabolas 
y^=z2px  and  x^  =  2py.  Ans.  f  pi 

12.  Find  the  total  area  included  between  the  curve  y  =  x^ 
and  the  line  y  =  2x.  Ans.  2. 

13.  Find  by  integration  the  area  of  the  triangle  bounded  by 
the  axis  of  Y and. the  lines  2x-\-y-{-S  =  0,  y  =  —  4:.       A^is.  4. 


216 


ELEMENTAKY  ANALYSIS 


14.  rind  the  area  bounded  by  each  of  the  following  curves 
and  the  X-axis. 

(a)  y  =  9  —  oc^. 

(b)  y  =  4:—x^. 

(c)  y=3  -2x-x\ 
■    (d)  y  =  5  —  4:  X  —  x^. 

(e)  2/  =  sin  X  +  cos  x  (one  arch). 
(/)  y  =  xV4.  —  x^. 
(g)  y  z=  x-\/9  —  x\ 

15.  Find  the  area  bounded  as  described. 


Ans.  36. 
Ans.  lOf. 
Ans.  lOf. 


Ans.  2-y/2, 
Ans,  2|. 


(^)  y- 


(P)  2/  = 
ip)  y 


,  ordinates  at  a;  =  0,  aj  =  3. 
ordinates  at  a:  =  0,  a;  =  3. 


l-{-x 
4 
(1  +  ^) 
tan  Xj  ordinates  at  a?  =  0,  a; 


IT 


Ans.  8  logg  2. 

Ans,  3. 

Ans,  ^  loge  2. 


(d)  y  =  V4  —  2  0?',  ordinates  at  aj  =  0,  a:^  =  2. 


^ns.  2|. 


^7is.  36. 
Ans,  \  IT. 


(e)  y  =  VlO  —  3x,  ordinates  at  oj  =  1,  a?  =  3. 

Ans. 
(/)  i»  =  9  —  ?/^,  and  the  ?/-axis. 
(^)  X  =  sin^?/?  ^^^  ^^^  2/-^xis. 

83.  Volumes  of  solids  of  revolutions.  Let  F denote  the  vol- 
ume of  the  solid  generated  by  revolving  the  plane  surface  ACDB 
about  the  axis  of  X,  the  equation  of  the  plane  curve  CPD  being 

(1)  y=f{x). 

Construct  rectangles  as  in  the 
figure.  When  the  area  is  re- 
volved each  rectangle  gener- 
ates a  cylinder  of  revolution. 
The  required  volume  is 
clearly  equal  to  the  limit  of 
X  the  sum  of  the  volumes  of 
these  cylinders. 


INTEGRATION   A   PROCESS   OF   SUMMATION      217 

Consider  any  one  of  the  cylinders,  say  the  one  generated  by 
the  rectangle  MNPQ.     Let  JO^=Aaj,  MP=y=f(x).     Then 


the  volume  of  the  cylinder  generated  is  iry^Ax  =  7rf(xy  Aa;,  and 
hence  the  expression  for  an  element  of  the  required  integral  is 

(2)  Try^Ax,  or  7rf(xyAx. 

The  Fundamental  Theorem  now  gives  the  formula 


(^) 


IT  (  p'^dx, 


where  the  value  of  y  in  terms  of  x  must  be  substituted  from 
the  equation  (1)  of  the  given  curve. 

This  formula  is  easily  recalled  if  we  consider  a  slice  or  disk 
of  the  solid  between  two  planes  perpendicular  to  the  axis  of 
revolution  as  an  element  of  the  volume,  and  regard  it  as  a 
cylinder  of  infinitesimal  altitude  dx  and  with  a  base  of  area 
7ry^,  —  hence  of  volume  iryHx.  Summing  up  all  such  slices 
(elements),  we  get  the  entire  volume.  Similarly,  when  OY 
is  the  axis  of  revolution,  we  use  the  formula 


(B) 


V=t^CxMy, 


where  the  value  of  x  in  terms  of  y  must  be  substituted  from 
the  equation  of  the  given  curve. 


218 


elemp:ntary  analysis 


EXAMPLE 
1.   Find   the   volume    generated   by   revolving    the   ellipse 


:  1  about  the  axis  of  X. 


Solution.     Since  if=  —  (a?  —  x^),  and  the  required  volume  is 
twice  the  volume  generated  by  OAB,  we  get,  substituting  in 


V 


y-dX  =  TT  I      -i:(ci^  —  X^)  dX  = 

Jo   or 


2  irab'^ 


V= 


4  iral 


To  verify  this  result,  let  b  ■• 


Then   F=^^,  the  vol- 


ume of  a  sphere,  which  is  only  a  special  case  of  the  ellipsoid. 
When  the  ellipse  is  revolved  about  its  major  axis  the  solid 
generated  is  called  a  prolate  spheroid ;  when  about  its  minor 
axis,  an  oblate  spheroid. 

Y 


PROBLEMS 

1.    Find  the  volume  of  the  sphere  generated  by  revolving 

4  TTT^ 


the  circle  x^  +  y^  =  r^  about  a  diameter. 


Ans. 


2.  Find  by  integration  the  volume  of  the  right  cone  gener- 
ated by  revolving  the  line  joining  the  origin  to  the  point  (a,  b) 
about  the  axis  of  X.     Verify  your  result  geometrically. 

Trab^ 
3 


Ans. 


INTEGRATION   A   PROCESS   OF   SUMMATION      219 

3.  Find  the  volume  of  the  cone  generated  by  revolving  the 
line  of  problem  2  about  the  axis  of  Y,  Verify  your  result  geo- 
metrically. Ans.  — -— . 

o 

4.  Find  the  volume  of  the  paraboloid  of  revolution  gener- 
ated by  revolving  the  arc  of  the  parabola  y^  =  4:ax  between 
the  origin  and  the  point  (xi,  y^)  about  its  axis. 

Ans.  2  irax^  =  ^^^  ^;  i.e.  one  half  of  the  volume  of  the  cir- 
cumscribing cylinder. 

5.  Find  the  volume  generated  by  revolving  the  arc  in  prob- 
lem 4  about  the  axis  of  Y, 

Ans,  ^^^^  =  -7ra?i^2/i5  ^•^-  ^"^^  M.^  of  the  cylinder  of  altitude 
80  a^     5 

?/i  and  radius  of  base  x^. 

6.  Find  by  integration  the  volume  of  the  cone  generated  by 
revolving   about    OX  that   part  of  the  line  4i»  —  5?/-f-3  =  0 

9  TT 

which  is  intercepted  between  the  coordinate  axes.       Ans. 

^  100 

7.  Find  the  volume  of  the  torus  (ring)  generated  by  revolv- 
ing the  circle  x^  +  (?/  —  lyf  —  a?  about  OX.  Ans.  2  7rW6. 

8.  Find  the  volume  generated  by  revolving  one  arch  of  the 

7J-2 

sine  curve  y  =  sin  x  about  the  axis  of  X.  Ans,   — 

^  2 

9.  Find  the  volumes  of  the  solids  obtained  by  revolving 
around  XX  each  of  the  areas  of  problem  14,  p.  216,  and  prob- 
lem 15  (a)-(e),  p.  216. 

84.  Miscellaneous  applications  of  the  Integral  Calculus. 

The  following  examples  illustrate  further  the  principle  of 
summation.  The  student  should  study  carefully  the  method 
of  subdividing  the  magnitude  to  be  calculated  into  elements, 
the  expression  for  an  element  ultimately  becoming  an  element 
of  the  required  integral  {{D),  p.  210). 


220  ELEMENTARY   ANALYSIS 

EXAMPLES 

1.  Determine  the  amount  of  attraction  exerted  by  a  thin, 
straight,  homogeneous  rod  of  uniform  thickness,  of  length  Z, 
and  of  mass  M,  upon  a  material  point  P  of  mass  m  situated  at 
a  distance  of  a  from  one  end  of  the  rod  in  its  line  of  direction. 


1 -.j 


Solution.     .Siippose  the  rod  to  be  divided  into  equal  infini- 
tesimal portions  (elements)  of  length  dx. 

M 

—  =  mass  of  a  unit  length  of  rod ; 

i 

M 
hence  —  dx=^  mass  of  any  element. 

Newton's   Law  for    measuring  the  attraction   between  any 
two  masses  is 

p  i?    i  J.      .  •  product  of  masses 

force  of  attraction  =  -—-^ :; ; 

(distance  between  them)'' 

therefore  the  force  of  attraction  between  the  particle  at  P  and 
an  element  of  the  rod  is 

—  mdx 
■  J 

-    (x  +  af' 

which  is  then  an  element  of  the  force  of  attraction  required. 
The  total  attraction  between  the  particle  at  P  and  the  rod 
being  the  sum  of  all  such  elements  between  a;  =  0  and  x=l, 
we  have 

—  mdx 
force  of  attraction :     * 


Jo  {x  +  aY 
Jo 


Mm  C^     dx  Mm  * 


I    Jq  {x-\-ay  a{a-\-T) 


INTEGRATION   A  PROCESS   OF   SUMMATION      221 


2.   A  heavy  cable  of  length  L  ft.  hangs  vertical, 
per  foot  is  w  lb.     Find  the  work   done   in   foot- 
pounds when  the  cable  is  hauled  up. 

Solution.  Draw  the  X-axis  from  the  upper  ex- 
tremity downward.  Consider  the  cable  as  made 
up  of  small  pieces  each  of  length  Ao;.  Then  the 
weight  of  each  of  these  pieces  is  w  ^x.  Consider 
any  one  of  these  pieces  whose  distance  from  the 
upper  end  of  the  cable  is  x.  Then  the  work  done 
in  hauling  up  this  piece  is  the  product  of  the 
weight  and  the  height  it  is  raised;  that  is, 

work  done  in  raising  one  piece  =  x  *  w  Ax. 

Hence  the  expression 

wx  Ax 


Ax 


is  an  element  of  the  required  integral. 

The  total  work  done  in  hauling  in  the  cable  is  therefore 
given  by  the  integral 


work  =   I     xwdx  =  w  I 

Jo  Jo 


xdx  =  - ft.  lb. 

2 


The  weight  of  the  cable  (=  W)  is  of  course  wL. 

.-.  work  done=  W^L) 

that  is,  the  work  done  is  equal  to  that  of  raising  a  weight  equal 
to  that  of  the  cable  a  height  equal  to  one  half  the  length  of  the 
cable. 

3.  A  perfect  gas  in  a  cylinder  expands  from  the  volume  Vq 
to  the  volume  v^,  the  temperature  remaining  constant.  Find 
the  work  done. 

Solution.  Let  the  original  level  (when  the  volume  is  Vq)  be 
the  circle  000,  and  the  final  level  (volume  =  Vi)  the  circle  QQQ* 
Consider  now  the  expansion  from  the  level  A  A  to  BB,  the 
height  AB  being  small. 


222 


ELEMENTARY   ANALYSIS 


Since  the  temperature  remains  constant,  the  pressure  and 
the  volume  during  the  expansion  satisfy 
Boyle's  Law, 

(1)  pv  =  constant  =  h. 

In  this  formula,  p  is  the  pressure  per 
unit  area. 

Now  the  work  done  when  the  gas  ex- 
pands from  the  level  AA  to  BB  is  the 
product  of  the  total  pressure  and  the 
height  AB, 

Let   the  cross   section   of  the  cylinder 

have  the  area  S.     Then  the  total  pressure 

equals  pS.     The  volume  between  the  levels 

AA  and  BB,  which  we  denote  by  Av,  is 

Av 


given  by  A^'  =  8  •  AB, 


AB  =  - 


Hence  the  element  of 


work  done,  that  is, 

(2)     the  'work  done  when  the  volume  increases  by  Av  =p8  •  AB 

=  p  Av. 

h 
This  must  be  expressed  in  terms  of  v.     By  (l),p  =  -,.-.  ele- 
ment of  work  done  =  A;  —  -      The  Fundamental  Theorem  ap- 

V 

plies  at  once,  and  gives 

roork  done  =  C'k  ^  =  k  C'-  =  A:  log  ^ .     Ans. 

Jv^  V  Jv.     V  Vo 


PROBLEMS 

1.  A  quantity  of  steam  expands  so  as  to  satisfy  the  law 
pv^-^^=C.  When  '^  =  1  cu.  ft.,  p  =  8000  lb.  per  square  foot. 
Find  the  work  done  in  expansion  from  'y  =  3  to  ?;  =  10. 

Ans.  7750  ft.  lb. 

2.  Find  the  work  done  in  the  expansion  of  a  quantity  of 
steam  from  2  cu.  ft.  at  a  pressure  of  2  T.  per  square  foot  to 
8  cu.  ft.     The  law  of  expansion  is  pv""-^  =  C.       Ans.  5.97  ft.  T. 


i:ntegration  a  process  of  summation    223 

3.  A  point  moves  along  a  line  so  that  the  relation  between 
the  velocity  and  time  is  one  of  the  following.  Find  the  dis- 
tance it  will  move  in  the  interval  of  time  indicated. 

(a)  v  =  lt',  t  =  0  to  t  =  2  sec.  Ans.  |. 

(b)  v  =  16  -  32  ^ ;  ^  =  0  to  ^  =  1  sec.  Ans.  0. 

(c)  v  =  S  cos  t'j  t  =  0  to  t  =  -.  Ans.  3. 

(d)  v  =  — — ;  ^  =  0to^  =  4.  Ans.  1.17. 

4.  A  sluiceway  is  closed  by  a  rectangular  gate  10  ft.  wide 
by  12  ft.  deep.  The  top  of  the  gate  is  level  with  the  water. - 
Find  the  pressure  against  the  gate,  assuming  that  1  cu.  ft.  of 
water  weighs  62  lb.  Ans.  44,640  lb. 

5.  Find  the  amount  of  the  attraction  in  Ex.  1,  Art.  84,  if 
the  material  point  P  is  directly  over  the  center  of  the  rod  and 
at  a  distance  a  from  it. 

Ans.  0,  where  6  is  the  angle  subtended  by  the  rod  for 

al 

an  observer  at  P. 


(-rt 


A 


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LD  21-100m-l,'54(1887sl6)476 


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